Samacheer Kalvi Class 11 Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Exercise 5.1

 

Question 1. Expand
(i) \( \left(2 x^{2}-\frac{3}{x}\right)^{3} \)
(ii) \( \left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4} \)
Answer:
(i) To expand \( \left(2 x^{2}-\frac{3}{x}\right)^{3} \), we use the binomial theorem \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).
Here, \( a = 2x^2 \) and \( b = -\frac{3}{x} \).
\( \left(2 x^{2}-\frac{3}{x}\right)^{3} = \binom{3}{0}(2x^2)^3\left(-\frac{3}{x}\right)^0 + \binom{3}{1}(2x^2)^2\left(-\frac{3}{x}\right)^1 + \binom{3}{2}(2x^2)^1\left(-\frac{3}{x}\right)^2 + \binom{3}{3}(2x^2)^0\left(-\frac{3}{x}\right)^3 \)
\( \implies = 1 \cdot (8x^6) \cdot 1 + 3 \cdot (4x^4) \cdot \left(-\frac{3}{x}\right) + 3 \cdot (2x^2) \cdot \left(\frac{9}{x^2}\right) + 1 \cdot 1 \cdot \left(-\frac{27}{x^3}\right) \)
\( \implies = 8x^6 - 36x^3 + 54 - \frac{27}{x^3} \)

(ii) To expand \( \left(2 x^{2}-3 \sqrt{1-x^{2}}\right)^{4}+\left(2 x^{2}+3 \sqrt{1-x^{2}}\right)^{4} \), let \( A = 2x^2 \) and \( B = 3\sqrt{1-x^2} \).
The expression becomes \( (A-B)^4 + (A+B)^4 \).
Using the binomial expansion:
\( (A+B)^4 = \binom{4}{0}A^4 + \binom{4}{1}A^3B + \binom{4}{2}A^2B^2 + \binom{4}{3}AB^3 + \binom{4}{4}B^4 \) ---(1)
\( (A-B)^4 = \binom{4}{0}A^4 - \binom{4}{1}A^3B + \binom{4}{2}A^2B^2 - \binom{4}{3}AB^3 + \binom{4}{4}B^4 \) ---(2)
Adding (1) and (2):
\( (A+B)^4 + (A-B)^4 = 2[\binom{4}{0}A^4 + \binom{4}{2}A^2B^2 + \binom{4}{4}B^4] \)
Now, substitute back \( A = 2x^2 \) and \( B = 3\sqrt{1-x^2} \).
\( = 2 \left[ 1 \cdot (2x^2)^4 + 6 \cdot (2x^2)^2 (3\sqrt{1-x^2})^2 + 1 \cdot (3\sqrt{1-x^2})^4 \right] \)
\( = 2 \left[ (16x^8) + 6 \cdot (4x^4) \cdot 9(1-x^2) + 81(1-x^2)^2 \right] \)
\( = 2 \left[ 16x^8 + 216x^4(1-x^2) + 81(1-x^2)^2 \right] \)
Expanding further:
\( = 2 \left[ 16x^8 + 216x^4 - 216x^6 + 81(1 - 2x^2 + x^4) \right] \)
\( = 2 \left[ 16x^8 - 216x^6 + 216x^4 + 81 - 162x^2 + 81x^4 \right] \)
\( = 2 \left[ 16x^8 - 216x^6 + (216+81)x^4 - 162x^2 + 81 \right] \)
\( = 2 \left[ 16x^8 - 216x^6 + 297x^4 - 162x^2 + 81 \right] \)
\( = 32x^8 - 432x^6 + 594x^4 - 324x^2 + 162 \)
In simple words: We used the binomial expansion formula to break down each part of the expression. For the first part, we applied the cube formula directly. For the second part, we noticed a pattern for (A+B) raised to a power plus (A-B) raised to the same power, which simplifies the calculation by cancelling out some terms.

🎯 Exam Tip: Remember the binomial theorem formulas for different powers, especially for \( (a+b)^n \) and its variations. Recognizing patterns like \( (A+B)^n + (A-B)^n \) can save significant calculation time in exams.

 

Question 2. Compute
(i) \( 102^4 \)
(ii) \( 99^4 \)
(iii) \( 9^7 \)
Answer:
(i) We can write \( 102^4 \) as \( (100+2)^4 \).
Using the binomial theorem \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \):
\( (100+2)^4 = \binom{4}{0}(100)^4(2)^0 + \binom{4}{1}(100)^3(2)^1 + \binom{4}{2}(100)^2(2)^2 + \binom{4}{3}(100)^1(2)^3 + \binom{4}{4}(100)^0(2)^4 \)
\( \implies = 1 \cdot (100,000,000) \cdot 1 + 4 \cdot (1,000,000) \cdot 2 + 6 \cdot (10,000) \cdot 4 + 4 \cdot (100) \cdot 8 + 1 \cdot 1 \cdot 16 \)
\( \implies = 100,000,000 + 8,000,000 + 240,000 + 3,200 + 16 \)
\( \implies = 108,243,216 \)

(ii) We can write \( 99^4 \) as \( (100-1)^4 \).
Using the binomial theorem \( (a-b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \):
\( (100-1)^4 = \binom{4}{0}(100)^4(-1)^0 + \binom{4}{1}(100)^3(-1)^1 + \binom{4}{2}(100)^2(-1)^2 + \binom{4}{3}(100)^1(-1)^3 + \binom{4}{4}(100)^0(-1)^4 \)
\( \implies = 1 \cdot (100,000,000) \cdot 1 + 4 \cdot (1,000,000) \cdot (-1) + 6 \cdot (10,000) \cdot 1 + 4 \cdot (100) \cdot (-1) + 1 \cdot 1 \cdot 1 \)
\( \implies = 100,000,000 - 4,000,000 + 60,000 - 400 + 1 \)
\( \implies = 96,059,601 \)

(iii) We can write \( 9^7 \) as \( (10-1)^7 \).
Using the binomial theorem:
\( (10-1)^7 = \binom{7}{0}(10)^7(-1)^0 + \binom{7}{1}(10)^6(-1)^1 + \binom{7}{2}(10)^5(-1)^2 + \binom{7}{3}(10)^4(-1)^3 + \binom{7}{4}(10)^3(-1)^4 + \binom{7}{5}(10)^2(-1)^5 + \binom{7}{6}(10)^1(-1)^6 + \binom{7}{7}(10)^0(-1)^7 \)
\( \implies = 1 \cdot 10^7 \cdot 1 + 7 \cdot 10^6 \cdot (-1) + 21 \cdot 10^5 \cdot 1 + 35 \cdot 10^4 \cdot (-1) + 35 \cdot 10^3 \cdot 1 + 21 \cdot 10^2 \cdot (-1) + 7 \cdot 10^1 \cdot 1 + 1 \cdot 1 \cdot (-1) \)
\( \implies = 10,000,000 - 7,000,000 + 2,100,000 - 350,000 + 35,000 - 2,100 + 70 - 1 \)
\( \implies = 4,782,969 \)
In simple words: To compute these powers, we rewrite the numbers as a sum or difference of a round number (like 100 or 10) and a small digit. Then, we use the binomial theorem to expand it, which helps us calculate the large numbers step-by-step. This makes it easier than multiplying the number by itself many times directly.

🎯 Exam Tip: When computing powers of numbers close to multiples of 10 or 100, always use the binomial theorem by writing them as \( (100+x)^n \) or \( (10-x)^n \). This simplifies calculations greatly and reduces errors compared to direct multiplication.

 

Question 3. Using binomial theorem, indicate which of the following two number is larger. \( (1.01)^{1000000}, 10000 \)
Answer:
Let's expand \( (1.01)^{1000000} \) using the binomial theorem. We can write \( 1.01 \) as \( (1 + 0.01) \).
So, \( (1 + 0.01)^{1000000} = \binom{1000000}{0}(1)^{1000000}(0.01)^0 + \binom{1000000}{1}(1)^{999999}(0.01)^1 + \binom{1000000}{2}(1)^{999998}(0.01)^2 + \dots \)
Let \( N = 1000000 \).
\( (1 + 0.01)^N = \binom{N}{0}(1)^N(0.01)^0 + \binom{N}{1}(1)^{N-1}(0.01)^1 + \text{other positive terms} \)
\( \implies = 1 \cdot 1 \cdot 1 + N \cdot 1 \cdot (0.01) + \text{other positive terms} \)
\( \implies = 1 + 1000000 \cdot (0.01) + \text{other positive terms} \)
\( \implies = 1 + 10000 + \text{other positive terms} \)
\( \implies = 10001 + \text{other positive terms} \)
Since the "other positive terms" are all greater than zero, we can conclude that:
\( (1.01)^{1000000} > 10001 \)
And we know that \( 10001 > 10000 \).
Therefore, \( (1.01)^{1000000} \) is larger than \( 10000 \).
In simple words: We used the binomial theorem to expand the number \( (1.01)^{1000000} \). We found that the first two terms of this expansion add up to 10001. Since all the other terms in the expansion are positive, the total value must be greater than 10001. Because 10001 is already larger than 10000, we can easily see that \( (1.01)^{1000000} \) is the bigger number.

🎯 Exam Tip: To compare values involving powers, especially when one is close to 1, using the binomial theorem's first few terms can quickly establish which value is larger without needing to calculate the exact number.

 

Question 4. Find the coefficient of \( x^{15} \) in \( \left(x^{2}+\frac{1}{x^{3}}\right)^{10} \)
Answer:
The general term \( T_{r+1} \) in the expansion of \( (a+b)^n \) is given by \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
In the given expansion \( \left(x^{2}+\frac{1}{x^{3}}\right)^{10} \), we have \( n=10 \), \( a=x^2 \), and \( b=\frac{1}{x^3} = x^{-3} \).
So, the general term is:
\( T_{r+1} = \binom{10}{r} (x^2)^{10-r} (x^{-3})^r \)
\( \implies = \binom{10}{r} x^{2(10-r)} x^{-3r} \)
\( \implies = \binom{10}{r} x^{20-2r-3r} \)
\( \implies = \binom{10}{r} x^{20-5r} \)
We want to find the coefficient of \( x^{15} \), so we set the exponent of \( x \) equal to 15:
\( 20 - 5r = 15 \)
\( \implies 5r = 20 - 15 \)
\( \implies 5r = 5 \)
\( \implies r = 1 \)
Now, substitute \( r=1 \) back into the general term formula to find the coefficient:
The term will be \( T_{1+1} = T_2 \).
\( T_2 = \binom{10}{1} x^{20-5(1)} \)
\( \implies T_2 = 10 x^{15} \)
Therefore, the coefficient of \( x^{15} \) is 10. The value of 'r' helps us identify the specific term. A binomial expansion creates terms in a series, and 'r' points to the position of the term we are looking for.
In simple words: First, we write down the general formula for any term in the expansion. Then, we find the power of \( x \) in that general term. We set this power equal to 15 (because we want the coefficient of \( x^{15} \)) and solve for 'r'. Once we have 'r', we put it back into the general term to find the number that multiplies \( x^{15} \), which is our coefficient.

🎯 Exam Tip: Always remember the formula for the general term \( T_{r+1} \) in binomial expansions. When finding a specific coefficient, equate the exponent of the variable in the general term to the required exponent and solve for \( r \).

 

Question 5. Find the coefficient of \( x^6 \) and the coefficient of \( x^2 \) in \( \left(x^{2}-\frac{1}{x^{3}}\right)^{6} \)
Answer:
The general term \( T_{r+1} \) in the expansion of \( (a+b)^n \) is given by \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
In the given expansion \( \left(x^{2}-\frac{1}{x^{3}}\right)^{6} \), we have \( n=6 \), \( a=x^2 \), and \( b=-\frac{1}{x^3} = -x^{-3} \).
So, the general term is:
\( T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-x^{-3})^r \)
\( \implies = \binom{6}{r} x^{2(6-r)} (-1)^r x^{-3r} \)
\( \implies = \binom{6}{r} (-1)^r x^{12-2r-3r} \)
\( \implies = \binom{6}{r} (-1)^r x^{12-5r} \)

**To find the coefficient of \( x^6 \):**
We set the exponent of \( x \) equal to 6:
\( 12 - 5r = 6 \)
\( \implies 5r = 12 - 6 \)
\( \implies 5r = 6 \)
\( \implies r = \frac{6}{5} \)
Since \( r \) must be a whole number (a non-negative integer), \( r = \frac{6}{5} \) is not possible. This indicates that there is no term with \( x^6 \) in the expansion.
Therefore, the coefficient of \( x^6 \) is 0. Knowing that 'r' must be a whole number is crucial when dealing with such problems.

**To find the coefficient of \( x^2 \):**
We set the exponent of \( x \) equal to 2:
\( 12 - 5r = 2 \)
\( \implies 5r = 12 - 2 \)
\( \implies 5r = 10 \)
\( \implies r = \frac{10}{5} = 2 \)
Now, substitute \( r=2 \) back into the general term formula to find the coefficient:
The term will be \( T_{2+1} = T_3 \).
\( T_3 = \binom{6}{2} (-1)^2 x^{12-5(2)} \)
\( \implies T_3 = \frac{6 \times 5}{2 \times 1} \cdot (1) \cdot x^{12-10} \)
\( \implies T_3 = 15 x^2 \)
Therefore, the coefficient of \( x^2 \) is 15.
In simple words: We first find the general form of any term in the expansion. For \( x^6 \), we set the power of \( x \) to 6 and found that 'r' was not a whole number, which means no \( x^6 \) term exists. For \( x^2 \), we set the power of \( x \) to 2 and found 'r' was 2. Using this 'r' value in the general term gave us the coefficient for \( x^2 \).

🎯 Exam Tip: Always check if the value of \( r \) (calculated by equating the variable's exponent) is a non-negative integer. If \( r \) is fractional or negative, it means the desired term does not exist in the expansion, and its coefficient is 0.

 

Question 6. Find the coefficient of \( x^4 \) in the expansion of \( (1 + x^3)^{50} \left(x^{2}+\frac{1}{x}\right)^{5} \)
Answer:
Let's find the general term for each part of the expression separately.

**Part 1: \( \left(x^{2}+\frac{1}{x}\right)^{5} \)**
The general term \( T_{r+1} \) for \( (a+b)^n \) is \( \binom{n}{r} a^{n-r} b^r \).
Here, \( n=5 \), \( a=x^2 \), \( b=\frac{1}{x} = x^{-1} \).
\( T_{r+1} = \binom{5}{r} (x^2)^{5-r} (x^{-1})^r \)
\( \implies = \binom{5}{r} x^{2(5-r)} x^{-r} \)
\( \implies = \binom{5}{r} x^{10-2r-r} \)
\( \implies = \binom{5}{r} x^{10-3r} \)
The terms in this expansion are:
For \( r=0 \): \( \binom{5}{0} x^{10} = 1x^{10} \)
For \( r=1 \): \( \binom{5}{1} x^7 = 5x^7 \)
For \( r=2 \): \( \binom{5}{2} x^4 = 10x^4 \)
For \( r=3 \): \( \binom{5}{3} x^1 = 10x \)
For \( r=4 \): \( \binom{5}{4} x^{-2} = 5x^{-2} \)
For \( r=5 \): \( \binom{5}{5} x^{-5} = 1x^{-5} \)
So, \( \left(x^{2}+\frac{1}{x}\right)^{5} = x^{10} + 5x^7 + 10x^4 + 10x + 5x^{-2} + x^{-5} \) ---(1)

**Part 2: \( (1 + x^3)^{50} \)**
The general term \( T_{k+1} \) is \( \binom{50}{k} (1)^{50-k} (x^3)^k \)
\( \implies = \binom{50}{k} x^{3k} \)
The terms in this expansion are:
For \( k=0 \): \( \binom{50}{0} x^0 = 1 \)
For \( k=1 \): \( \binom{50}{1} x^3 = 50x^3 \)
For \( k=2 \): \( \binom{50}{2} x^6 = \frac{50 \times 49}{2} x^6 = 1225x^6 \)
For \( k=3 \): \( \binom{50}{3} x^9 = \frac{50 \times 49 \times 48}{3 \times 2 \times 1} x^9 = 19600x^9 \)
So, \( (1 + x^3)^{50} = 1 + 50x^3 + 1225x^6 + 19600x^9 + \dots \) ---(2)

Now we need to find the coefficient of \( x^4 \) in the product of (1) and (2).
We look for pairs of terms, one from each expansion, whose exponents of \( x \) add up to 4.
From (1) and (2):
\( (10x^4) \cdot (1) = 10x^4 \)
\( (x^1 \text{ term, which is } 10x) \cdot (\text{no } x^3 \text{ term in } (1+x^3)^{50} \text{ for } k=1 \text{ from } x^{3k}) \) *Correction: This is not correct. Need to find terms such that powers add to 4.*
Let's list the terms from (1) and (2) whose product can give \( x^4 \):
1. Term from (1) with \( x^4 \): \( 10x^4 \). Multiplied by constant term from (2): \( 1 \). Product: \( 10x^4 \).
2. Term from (1) with \( x^1 \): \( 10x \). We need \( x^3 \) from (2). Term from (2) with \( x^3 \): \( 50x^3 \). Product: \( 10x \cdot 50x^3 = 500x^4 \).
3. Term from (1) with \( x^7 \): \( 5x^7 \). We need \( x^{-3} \) from (2). No \( x^{-3} \) term in (2).
4. Term from (1) with \( x^{10} \): \( 1x^{10} \). We need \( x^{-6} \) from (2). No \( x^{-6} \) term in (2).
5. Term from (1) with \( x^{-2} \): \( 5x^{-2} \). We need \( x^6 \) from (2). Term from (2) with \( x^6 \): \( 1225x^6 \). Product: \( 5x^{-2} \cdot 1225x^6 = 6125x^4 \).
6. Term from (1) with \( x^{-5} \): \( 1x^{-5} \). We need \( x^9 \) from (2). Term from (2) with \( x^9 \): \( 19600x^9 \). Product: \( 1x^{-5} \cdot 19600x^9 = 19600x^4 \).

Summing the coefficients of \( x^4 \):
Total coefficient of \( x^4 = 10 + 500 + 6125 + 19600 \)
\( = 26235 \)
Therefore, the coefficient of \( x^4 \) is 26235. To successfully combine expansions, you must ensure that all relevant term pairs are identified and their coefficients are summed correctly.
In simple words: We first found all the terms for each part of the expression. Then, we looked for pairs of terms, one from each part, whose \( x \) powers would add up to 4. For each such pair, we multiplied their coefficients. Finally, we added up all these resulting coefficients to get the total coefficient for \( x^4 \).

🎯 Exam Tip: When finding a coefficient in a product of two binomial expansions, list out the significant terms for each expansion and then systematically identify all pairs of terms whose product yields the desired power of \( x \).

 

Question 7. Find the constant term of \( \left(x^{3}-\frac{1}{3 x^{2}}\right)^{5} \)
Answer:
The general term \( T_{r+1} \) in the expansion of \( (a+b)^n \) is given by \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
In the given expansion \( \left(x^{3}-\frac{1}{3 x^{2}}\right)^{5} \), we have \( n=5 \), \( a=x^3 \), and \( b=-\frac{1}{3x^2} = -\frac{1}{3}x^{-2} \).
So, the general term is:
\( T_{r+1} = \binom{5}{r} (x^3)^{5-r} \left(-\frac{1}{3}x^{-2}\right)^r \)
\( \implies = \binom{5}{r} x^{3(5-r)} \left(-\frac{1}{3}\right)^r x^{-2r} \)
\( \implies = \binom{5}{r} \left(-\frac{1}{3}\right)^r x^{15-3r-2r} \)
\( \implies = \binom{5}{r} \left(-\frac{1}{3}\right)^r x^{15-5r} \)
For the term to be a constant term, the exponent of \( x \) must be 0.
So, we set the exponent of \( x \) equal to 0:
\( 15 - 5r = 0 \)
\( \implies 5r = 15 \)
\( \implies r = 3 \)
Now, substitute \( r=3 \) back into the general term formula to find the constant term:
The term will be \( T_{3+1} = T_4 \).
\( T_4 = \binom{5}{3} \left(-\frac{1}{3}\right)^3 x^{15-5(3)} \)
\( \implies T_4 = \frac{5 \times 4}{2 \times 1} \cdot \left(-\frac{1}{27}\right) \cdot x^0 \)
\( \implies T_4 = 10 \cdot \left(-\frac{1}{27}\right) \cdot 1 \)
\( \implies T_4 = -\frac{10}{27} \)
Therefore, the constant term is \( -\frac{10}{27} \). The constant term is a term that does not contain any variable, hence its exponent is zero.
In simple words: We used the general term formula for binomial expansion. We then set the power of \( x \) in this formula to zero, because a constant term doesn't have \( x \). Solving for 'r' gave us the position of the constant term. Finally, we put 'r' back into the formula to find the actual value of that constant term.

🎯 Exam Tip: To find the constant term in a binomial expansion, always set the exponent of the variable in the general term to zero. This helps you determine the value of \( r \) for that specific term.

 

Question 8. Find the last two digits of the number \( 3^{600} \).
Answer:
To find the last two digits of a number, we need to find its value modulo 100.
We have \( 3^{600} = (3^2)^{300} = 9^{300} \).
We can write \( 9^{300} \) as \( (10-1)^{300} \).
Using the binomial theorem for \( (a-b)^n \):
\( (10-1)^{300} = \binom{300}{0}(10)^{300}(-1)^0 + \binom{300}{1}(10)^{299}(-1)^1 + \dots + \binom{300}{298}(10)^2(-1)^{298} + \binom{300}{299}(10)^1(-1)^{299} + \binom{300}{300}(10)^0(-1)^{300} \)
We are interested in the last two digits, which means we care about terms that affect the value modulo 100. Any term with \( (10)^2 \) or higher power of 10 will be a multiple of 100, and thus will not affect the last two digits.
So, we only need to consider the last two terms in the expansion:
\( \binom{300}{299}(10)^1(-1)^{299} + \binom{300}{300}(10)^0(-1)^{300} \)
\( \implies = 300 \cdot 10 \cdot (-1) + 1 \cdot 1 \cdot 1 \)
\( \implies = -3000 + 1 \)
\( \implies = -2999 \)
To find the last two digits, we take \( -2999 \pmod{100} \).
\( -2999 = -3000 + 1 \). When divided by 100, \( -3000 \) leaves a remainder of 0.
So, \( -2999 \pmod{100} = 1 \pmod{100} \).
Alternatively, \( -2999 + 3000 = 1 \). Since \( 3000 \) is a multiple of 100, the remainder is 1.
Thus, the last two digits of \( 3^{600} \) are 01.
In simple words: We changed \( 3^{600} \) into \( (10-1)^{300} \). Then, we used the binomial theorem but only looked at the last few terms. This is because all the terms with \( 10^2 \) (which is 100) or higher powers of 10 will end in 00. So, only the terms with \( 10^1 \) and \( 10^0 \) affect the last two digits. Calculating these terms gave us -2999, which has 01 as its last two digits.

🎯 Exam Tip: To find the last few digits of a large power, express the base as \( (10 \pm x)^n \) or \( (100 \pm x)^n \) and use the binomial theorem. Only the terms with powers of 10 less than the number of digits required (e.g., \( 10^1 \) for last two digits) will contribute to the answer.

 

Question 9. If n is a positive integer show that \( 9^{n+1} – 8n – 9 \) is always divisible by 64.
Answer:
We can rewrite \( 9^{n+1} \) as \( (1+8)^{n+1} \).
Using the binomial theorem for \( (1+x)^k = \binom{k}{0} + \binom{k}{1}x + \binom{k}{2}x^2 + \binom{k}{3}x^3 + \dots + \binom{k}{k}x^k \).
Let \( k = n+1 \) and \( x = 8 \).
\( (1+8)^{n+1} = \binom{n+1}{0} + \binom{n+1}{1}(8) + \binom{n+1}{2}(8)^2 + \binom{n+1}{3}(8)^3 + \dots + \binom{n+1}{n+1}(8)^{n+1} \)
\( \implies 9^{n+1} = 1 + (n+1)8 + \frac{(n+1)n}{2 \times 1}8^2 + \text{terms with } 8^3 \text{ or higher powers of 8} \)
\( \implies 9^{n+1} = 1 + 8(n+1) + \frac{n(n+1)}{2} \times 64 + \text{terms divisible by } 8^3 \text{ (which is 512)} \)
\( \implies 9^{n+1} = 1 + 8n + 8 + 32n(n+1) + \text{terms divisible by 512} \)
\( \implies 9^{n+1} = 9 + 8n + 32n(n+1) + \text{terms divisible by 512} \)
Now, let's rearrange the given expression \( 9^{n+1} – 8n – 9 \):
\( 9^{n+1} – 8n – 9 = (9 + 8n + 32n(n+1) + \text{terms divisible by 512}) – 8n – 9 \)
\( \implies 9^{n+1} – 8n – 9 = 32n(n+1) + \text{terms divisible by 512} \)
Since \( 512 \) is a multiple of \( 64 \) (i.e., \( 512 = 64 \times 8 \)), the terms divisible by 512 are also divisible by 64.
We need to show that \( 32n(n+1) \) is divisible by 64.
The product \( n(n+1) \) is always an even number, because either \( n \) or \( n+1 \) is even. Let \( n(n+1) = 2m \) for some integer \( m \).
Then \( 32n(n+1) = 32(2m) = 64m \).
Since \( 64m \) is a multiple of 64, it is divisible by 64.
Therefore, \( 9^{n+1} – 8n – 9 \) is the sum of terms, all of which are divisible by 64.
This proves that \( 9^{n+1} – 8n – 9 \) is always divisible by 64 for any positive integer \( n \). The divisibility of \( n(n+1) \) by 2 is a key observation that simplifies this proof.
In simple words: We changed \( 9^{n+1} \) to \( (1+8)^{n+1} \) and expanded it using the binomial theorem. We kept only the first few terms and grouped the rest, showing they are divisible by large powers of 8. We then showed that the remaining part of the expression is also a multiple of 64. Since all parts are divisible by 64, the whole expression must be divisible by 64.

🎯 Exam Tip: When proving divisibility using binomial expansion, express the base number as \( (1 \pm x)^n \). Expand only enough terms to isolate a multiple of the divisor, then prove the divisibility of the remaining terms.

 

Question 10. If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of \( (x + y)^n \) are equal.
Answer:
When \( n \) is an odd positive integer, the binomial expansion of \( (x+y)^n \) has \( n+1 \) terms. Since \( n \) is odd, \( n+1 \) is even. Therefore, there will be two middle terms.
The position of the middle terms are \( \frac{n+1}{2} \) and \( \frac{n+1}{2} + 1 \).

Let's find the coefficients of these terms.
The \( T_{r+1} \) term in the expansion of \( (x+y)^n \) is \( \binom{n}{r} x^{n-r} y^r \). The coefficient is \( \binom{n}{r} \).

**First middle term:**
The position is \( \frac{n+1}{2} \). So, \( r+1 = \frac{n+1}{2} \).
\( \implies r = \frac{n+1}{2} - 1 = \frac{n+1-2}{2} = \frac{n-1}{2} \).
The coefficient of the first middle term is \( \binom{n}{\frac{n-1}{2}} \).

**Second middle term:**
The position is \( \frac{n+1}{2} + 1 \). So, \( r+1 = \frac{n+1}{2} + 1 \).
\( \implies r = \frac{n+1}{2} \).
The coefficient of the second middle term is \( \binom{n}{\frac{n+1}{2}} \).

Now, we need to prove that \( \binom{n}{\frac{n-1}{2}} = \binom{n}{\frac{n+1}{2}} \).
We know the property of binomial coefficients: \( \binom{n}{k} = \binom{n}{n-k} \).
Let's apply this property to the first coefficient, \( \binom{n}{\frac{n-1}{2}} \):
\( \binom{n}{\frac{n-1}{2}} = \binom{n}{n - \frac{n-1}{2}} \)
\( \implies = \binom{n}{\frac{2n - (n-1)}{2}} \)
\( \implies = \binom{n}{\frac{2n - n + 1}{2}} \)
\( \implies = \binom{n}{\frac{n+1}{2}} \)
This shows that the coefficient of the first middle term is equal to the coefficient of the second middle term.
Thus, the coefficients of the middle terms in the expansion of \( (x+y)^n \) are equal when \( n \) is an odd positive integer. Understanding how \( n \) being odd affects the number and position of middle terms is crucial here.
In simple words: When 'n' is an odd number, there are two middle terms in the expansion. We found the 'r' value for each of these middle terms. Then, using a property of binomial coefficients (that \( \binom{n}{k} \) is the same as \( \binom{n}{n-k} \)), we showed that the coefficient for the first middle term is exactly the same as the coefficient for the second middle term.

🎯 Exam Tip: Remember that for an odd \( n \), there are two middle terms in a binomial expansion, located at positions \( \frac{n+1}{2} \) and \( \frac{n+3}{2} \) (or \( \frac{n+1}{2} + 1 \)). The key property \( \binom{n}{k} = \binom{n}{n-k} \) is essential for proving the equality of their coefficients.

 

Question 11. If n is a positive integer and r is a non-negative integer, prove that the coefficients of \( x^r \) and \( x^{n-r} \) in the expansion of \( (1 + x)^n \) are equal.
Answer:
The general term \( T_{k+1} \) in the expansion of \( (1+x)^n \) is given by \( T_{k+1} = \binom{n}{k} (1)^{n-k} (x)^k = \binom{n}{k} x^k \).

**Coefficient of \( x^r \):**
To find the coefficient of \( x^r \), we set \( k=r \).
So, the term containing \( x^r \) is \( T_{r+1} = \binom{n}{r} x^r \).
The coefficient of \( x^r \) is \( \binom{n}{r} \). ---(1)

**Coefficient of \( x^{n-r} \):**
To find the coefficient of \( x^{n-r} \), we set \( k=n-r \).
So, the term containing \( x^{n-r} \) is \( T_{(n-r)+1} = \binom{n}{n-r} x^{n-r} \).
The coefficient of \( x^{n-r} \) is \( \binom{n}{n-r} \). ---(2)

Now, we need to prove that \( \binom{n}{r} = \binom{n}{n-r} \).
This is a fundamental property of binomial coefficients, often stated as \( \text{nCr} = \text{nC(n-r)} \).
Let's prove it:
\( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
And \( \binom{n}{n-r} = \frac{n!}{(n-r)!(n-(n-r))!} = \frac{n!}{(n-r)!(n-n+r)!} = \frac{n!}{(n-r)!r!} \)
Since \( r!(n-r)! = (n-r)!r! \), we can see that \( \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!} \).
Therefore, \( \binom{n}{r} = \binom{n}{n-r} \).
This proves that the coefficients of \( x^r \) and \( x^{n-r} \) in the expansion of \( (1+x)^n \) are equal. The symmetric nature of binomial coefficients around the middle term is a key concept here.
In simple words: We found the coefficient of \( x^r \) using the general term formula. Then, we found the coefficient of \( x^{n-r} \) in the same way. We then showed that these two coefficients are always the same by using a basic rule for combinations: choosing 'r' items from 'n' is the same as choosing 'n-r' items from 'n'.

🎯 Exam Tip: Always remember the symmetry property of binomial coefficients, \( \binom{n}{r} = \binom{n}{n-r} \). This property is frequently used in proofs and simplifying calculations related to coefficients in binomial expansions.

 

Question 12. If a and b are distinct integers, prove that \( a - b \) is a factor of \( a^n – b^n \), whenever \( n \) is a positive integer.
Answer:
We need to show that \( a^n - b^n \) is divisible by \( a-b \).
Let's write \( a \) as \( (b + (a-b)) \).
So, \( a^n = (b + (a-b))^n \).
Using the binomial theorem for \( (X+Y)^n = \binom{n}{0}X^n + \binom{n}{1}X^{n-1}Y + \binom{n}{2}X^{n-2}Y^2 + \dots + \binom{n}{n-1}XY^{n-1} + \binom{n}{n}Y^n \).
Let \( X=b \) and \( Y=(a-b) \).
\( a^n = \binom{n}{0}b^n + \binom{n}{1}b^{n-1}(a-b) + \binom{n}{2}b^{n-2}(a-b)^2 + \dots + \binom{n}{n-1}b(a-b)^{n-1} + \binom{n}{n}(a-b)^n \)
\( \implies a^n = b^n + \binom{n}{1}b^{n-1}(a-b) + \binom{n}{2}b^{n-2}(a-b)^2 + \dots + \binom{n}{n-1}b(a-b)^{n-1} + (a-b)^n \)
Now, subtract \( b^n \) from both sides:
\( a^n - b^n = \binom{n}{1}b^{n-1}(a-b) + \binom{n}{2}b^{n-2}(a-b)^2 + \dots + \binom{n}{n-1}b(a-b)^{n-1} + (a-b)^n \)
Notice that every term on the right-hand side has \( (a-b) \) as a factor. We can factor out \( (a-b) \):
\( a^n - b^n = (a-b) \left[ \binom{n}{1}b^{n-1} + \binom{n}{2}b^{n-2}(a-b) + \dots + \binom{n}{n-1}b(a-b)^{n-2} + (a-b)^{n-1} \right] \)
Since \( a^n - b^n \) can be expressed as \( (a-b) \) multiplied by another integer (the expression in the square brackets), it means that \( (a-b) \) is a factor of \( a^n - b^n \). This identity is useful in algebra for simplifying expressions.
In simple words: We rewrote 'a' as 'b plus (a-b)'. Then we used the binomial theorem to expand \( a^n \). After subtracting \( b^n \) from both sides, we noticed that every single term on the right side had '(a-b)' as a common multiplier. This shows that we can factor out '(a-b)', proving that it is a factor of \( a^n - b^n \).

🎯 Exam Tip: To prove that \( (a-b) \) is a factor of \( (a^n-b^n) \), express \( a \) as \( (b + (a-b)) \) and use the binomial theorem. After expansion and rearranging, you should be able to factor out \( (a-b) \) from the expression.

 

Question 13. In the binomial expansion of \( (a + b)^n \), the coefficients of the 4th and 13th terms are equal to each other, find n.
Answer:
The general term \( T_{r+1} \) in the expansion of \( (a+b)^n \) is \( \binom{n}{r} a^{n-r} b^r \). The coefficient of this term is \( \binom{n}{r} \).

For the 4th term: \( r+1 = 4 \implies r = 3 \).
The coefficient of the 4th term is \( \binom{n}{3} \).

For the 13th term: \( r+1 = 13 \implies r = 12 \).
The coefficient of the 13th term is \( \binom{n}{12} \).

We are given that the coefficients of the 4th and 13th terms are equal:
\( \binom{n}{3} = \binom{n}{12} \)
We know that if \( \binom{n}{x} = \binom{n}{y} \), then either \( x = y \) or \( x + y = n \).
Since \( 3 \neq 12 \), we must have \( x+y=n \).
So, \( 3 + 12 = n \).
\( \implies n = 15 \)
Therefore, the value of \( n \) is 15. This property of binomial coefficients helps quickly solve for 'n' without complex calculations.
In simple words: The problem says that the numbers in front of the 4th term and the 13th term in the expansion are the same. We know a rule that if two combinations \( \binom{n}{x} \) and \( \binom{n}{y} \) are equal, then either \( x \) must be equal to \( y \), or \( x+y \) must be equal to \( n \). Since 4 is not 13, we used the second rule: \( 4+13 \) equals \( n \), which gives us \( n=15 \).

🎯 Exam Tip: When given that two coefficients \( \binom{n}{x} \) and \( \binom{n}{y} \) are equal, immediately apply the property: if \( x \neq y \), then \( x+y=n \). This is a common shortcut in binomial theorem problems.

 

Question 14. If the coefficients of three consecutive terms in the expansion of \( (a + x)^n \) are in the ratio 1 : 7 : 42, then find n.
Answer:
Let the three consecutive terms be \( T_r \), \( T_{r+1} \), and \( T_{r+2} \).
The general term \( T_{k+1} \) in the expansion of \( (a+x)^n \) is \( \binom{n}{k} a^{n-k} x^k \). The coefficient is \( \binom{n}{k} \).

Coefficient of \( T_r \): This corresponds to \( k = r-1 \). So, \( \binom{n}{r-1} \).
Coefficient of \( T_{r+1} \): This corresponds to \( k = r \). So, \( \binom{n}{r} \).
Coefficient of \( T_{r+2} \): This corresponds to \( k = r+1 \). So, \( \binom{n}{r+1} \).

We are given that these coefficients are in the ratio 1 : 7 : 42.
So, \( \binom{n}{r-1} : \binom{n}{r} = 1 : 7 \) and \( \binom{n}{r} : \binom{n}{r+1} = 7 : 42 \).

**From the first ratio:**
\( \frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{1}{7} \)
We know the identity \( \frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1} \). So, let \( k=r \).
\( \frac{r}{n-r+1} = \frac{1}{7} \)
\( \implies 7r = n-r+1 \)
\( \implies 8r = n+1 \) ---(1)

**From the second ratio:**
\( \frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{7}{42} = \frac{1}{6} \)
We know the identity \( \frac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k} \). So, let \( k=r \).
\( \frac{r+1}{n-r} = \frac{1}{6} \)
\( \implies 6(r+1) = n-r \)
\( \implies 6r+6 = n-r \)
\( \implies 7r+6 = n \) ---(2)

Now we have a system of two linear equations with two variables \( n \) and \( r \):
1. \( n - 8r = -1 \)
2. \( n - 7r = 6 \)

Subtract equation (1) from equation (2):
\( (n - 7r) - (n - 8r) = 6 - (-1) \)
\( \implies n - 7r - n + 8r = 6 + 1 \)
\( \implies r = 7 \)

Substitute the value of \( r=7 \) into equation (2):
\( n - 7(7) = 6 \)
\( \implies n - 49 = 6 \)
\( \implies n = 6 + 49 \)
\( \implies n = 55 \)
Therefore, the value of \( n \) is 55. The use of ratios of consecutive terms is a powerful technique for solving such problems.
In simple words: We chose three general consecutive terms and wrote down their coefficients. We used the given ratios to set up two equations involving 'n' and 'r'. We solved these two equations together to find 'r' first, and then used 'r' to find 'n'. This method helps us find 'n' without having to write out the full binomial expansion.

🎯 Exam Tip: When dealing with ratios of consecutive binomial coefficients, use the handy identities \( \frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1} \) and \( \frac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k} \). This will simplify setting up and solving the equations for \( n \) and \( r \).

 

Question 15. In the binomial coefficients of \( (1 + x)^n \), the coefficients of the 5th, 6th, and 7th terms are in A.P. Find all values of n.
Answer:
The general term \( T_{k+1} \) in the expansion of \( (1+x)^n \) is \( \binom{n}{k} (1)^{n-k} (x)^k = \binom{n}{k} x^k \). The coefficient is \( \binom{n}{k} \).

For the 5th term: \( k+1 = 5 \implies k = 4 \). Coefficient is \( \binom{n}{4} \).
For the 6th term: \( k+1 = 6 \implies k = 5 \). Coefficient is \( \binom{n}{5} \).
For the 7th term: \( k+1 = 7 \implies k = 6 \). Coefficient is \( \binom{n}{6} \).

We are given that these coefficients are in Arithmetic Progression (A.P.).
If \( A, B, C \) are in A.P., then \( 2B = A+C \).
So, \( 2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6} \).

Let's write out the combinations:
\( 2 \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!} \)
Divide all terms by \( n! \):
\( \frac{2}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!} \)
We know that \( 5! = 5 \cdot 4! \), \( 6! = 6 \cdot 5 \cdot 4! \), \( (n-4)! = (n-4)(n-5)! \), and \( (n-5)! = (n-5)(n-6)! \).

\( \frac{2}{5 \cdot 4! (n-5)(n-6)!} = \frac{1}{4! (n-4)(n-5)(n-6)!} + \frac{1}{6 \cdot 5 \cdot 4! (n-6)!} \)
Multiply the entire equation by \( 4! (n-6)! \):
\( \frac{2}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30} \)
Multiply by \( 30(n-4)(n-5) \) to clear denominators:
\( 2 \cdot 6 (n-4) = 30 + (n-4)(n-5) \)
\( 12(n-4) = 30 + n^2 - 5n - 4n + 20 \)
\( 12n - 48 = 30 + n^2 - 9n + 20 \)
\( 12n - 48 = n^2 - 9n + 50 \)
Rearrange into a quadratic equation:
\( n^2 - 9n - 12n + 50 + 48 = 0 \)
\( n^2 - 21n + 98 = 0 \)
Factor the quadratic equation:
We need two numbers that multiply to 98 and add to -21. These are -7 and -14.
\( (n-7)(n-14) = 0 \)
So, \( n-7=0 \implies n=7 \)
Or \( n-14=0 \implies n=14 \)

Both values are valid since \( n \) must be at least 6 for the 7th term to exist. Understanding the properties of Arithmetic Progression is crucial here.
Therefore, the possible values of \( n \) are 7 and 14.
In simple words: We wrote down the coefficients for the 5th, 6th, and 7th terms using the combination formula. Since they are in an Arithmetic Progression, the middle coefficient (6th term) multiplied by 2 must equal the sum of the other two coefficients (5th and 7th terms). We then solved this equation for 'n' by simplifying the combination terms, which gave us a quadratic equation. Solving the quadratic equation provided the two possible values for 'n'.

🎯 Exam Tip: When coefficients of terms are in A.P., use the property \( 2B = A+C \). Simplify the combinations by cancelling out common factorials to arrive at a quadratic equation, which will give the possible values for \( n \). Always check that \( n \) is large enough for all terms to exist.

 

Question 16. Prove that \( \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2 = \frac{(2n)!}{(n!)^2} \)
Answer:
We need to prove that \( \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \). This is a standard identity known as Vandermonde's Identity for a specific case, or the Hockey-stick Identity if viewed differently.

Consider the expansion of \( (1+x)^n \):
\( (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n \) ---(1)

Now, consider the expansion of \( (x+1)^n \). Using the property \( \binom{n}{k} = \binom{n}{n-k} \), we can write:
\( (x+1)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \binom{n}{2}x^{n-2} + \dots + \binom{n}{n}x^0 \) ---(2)

Now, let's multiply these two expansions:
\( (1+x)^n \cdot (x+1)^n = (1+x)^{2n} \)
The coefficient of \( x^n \) in the expansion of \( (1+x)^{2n} \) is \( \binom{2n}{n} \).

Let's also find the coefficient of \( x^n \) by multiplying the series (1) and (2):
The coefficient of \( x^n \) in the product of \( (\binom{n}{0} + \binom{n}{1}x + \dots + \binom{n}{n}x^n) \) and \( (\binom{n}{0}x^n + \binom{n}{1}x^{n-1} + \dots + \binom{n}{n}x^0) \) is found by pairing terms such that their powers of \( x \) sum to \( n \):
\( (\binom{n}{0} \cdot \binom{n}{0}x^n) + (\binom{n}{1}x \cdot \binom{n}{1}x^{n-1}) + (\binom{n}{2}x^2 \cdot \binom{n}{2}x^{n-2}) + \dots + (\binom{n}{n}x^n \cdot \binom{n}{n}x^0) \)
\( \implies \binom{n}{0}^2 x^n + \binom{n}{1}^2 x^n + \binom{n}{2}^2 x^n + \dots + \binom{n}{n}^2 x^n \)
So, the coefficient of \( x^n \) in the product is \( \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2 \).

Since the coefficient of \( x^n \) in \( (1+x)^{2n} \) must be equal to the coefficient of \( x^n \) in the product of the two series, we have:
\( \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2 = \binom{2n}{n} \)
And we know that \( \binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2} \).
Thus, \( \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2 = \frac{(2n)!}{(n!)^2} \). This proof beautifully connects the expansion of polynomials with combinatorial identities.
In simple words: We considered the expansion of \( (1+x)^n \) multiplied by itself, which gives \( (1+x)^{2n} \). We know the coefficient of \( x^n \) in \( (1+x)^{2n} \) is \( \binom{2n}{n} \). We also found the coefficient of \( x^n \) by multiplying the individual terms of \( (1+x)^n \) with \( (x+1)^n \) (using the reverse order of powers). This led to a sum of squared coefficients. Since both methods find the same coefficient, they must be equal, proving the identity.

🎯 Exam Tip: To prove identities involving sums of squared binomial coefficients, consider the product of two binomial expansions like \( (1+x)^n \) and \( (x+1)^n \). Then, compare the coefficient of a specific power of \( x \) (often \( x^n \)) from both sides of the equation.

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