GSEB Class 11 Maths Solutions Chapter 8 Binomial Theorem Exercise 8.2

Question 1. Find the coefficient of x⁵ in \( (x + 3)^8 \)
Answer: The general term in the expansion of \( (x + 3)^8 \) is given by \( T_{r+1} = \binom{8}{r} x^{8-r} 3^r \). We need to find the coefficient of \( x^5 \). To do this, we set the exponent of x equal to 5:
\( 8 - r = 5 \)
\( r = 8 - 5 \)
\( r = 3 \)
Now, we substitute \( r = 3 \) back into the general term to find the coefficient:
Coefficient of \( x^5 = \binom{8}{3} 3^3 \)
\( = \frac{8 \times 7 \times 6}{1 \times 2 \times 3} \times 27 \)
\( = 56 \times 27 \)
\( = 1512 \)
In simple words: To find the number multiplying \( x^5 \), we use the general term formula for the binomial expansion. We figure out which 'r' value gives \( x^5 \), then plug that 'r' into the formula to calculate the coefficient.

Exam Tip: Remember the general term formula \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \). Clearly identify 'a' and 'b' from the given expression, and set the exponent of the required term to find 'r'.

Question 2. Find the coefficient of a⁵b⁷ in \( (a – 2b)^{12} \)
Answer: The given binomial expression is \( (a – 2b)^{12} \). We can write this as \( [a + (-2b)]^{12} \).
The general term \( T_{r+1} \) in this expansion is \( \binom{12}{r} a^{12-r} (-2b)^r \).
This can be written as \( \binom{12}{r} a^{12-r} (-2)^r b^r \).
We need to find the coefficient of \( a^5 b^7 \). So, we compare the powers:
For \( a \): \( 12 - r = 5 \implies r = 7 \)
For \( b \): \( r = 7 \)
Since both conditions give \( r = 7 \), we can substitute this value into the general term:
\( T_{7+1} = \binom{12}{7} a^{12-7} (-2b)^7 \)
\( = \binom{12}{7} a^5 (-2)^7 b^7 \)
The required coefficient is \( \binom{12}{7} (-2)^7 \).
\( \binom{12}{7} = \frac{12!}{7!(12-7)!} = \frac{12!}{7!5!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( = 12 \times 11 \times 3 \times 2 / (5 \times 4 \times 3 \times 2 \times 1) \) -> \( = 12 \times 11 \times 10 \times 9 \times 8 / (5 \times 4 \times 3 \times 2 \times 1) \)
\( = 12 \times 11 \times 10 \times 9 \times 8 / 120 \)
\( = 792 \)
\( (-2)^7 = -128 \)
So, the coefficient is \( 792 \times (-128) = -101376 \).
In simple words: First, we write the general term for the binomial. Then, we match the powers of 'a' and 'b' to find 'r'. Once we have 'r', we plug it back in to calculate the number that goes with \( a^5 b^7 \), including the negative sign.

Exam Tip: Pay careful attention to negative signs in the 'b' term (e.g., \( -2b \)) and ensure they are raised to the correct power 'r', as this often changes the final sign of the coefficient.

Question 3. General term = \( T_{r+1} = \binom{6}{r} (x^2)^{6-r}(- y)^r \)
Answer: The given general term is for the expansion of \( (x^2 - y)^6 \). Let's expand and simplify this general term:
\( T_{r+1} = \binom{6}{r} (x^2)^{6-r}(- y)^r \)
\( = \binom{6}{r} x^{2(6-r)} (-1)^r y^r \)
\( = (-1)^r \frac{6!}{r!(6 - r)!} x^{12-2r} y^r \)
This is the simplified form of the general term.
In simple words: This problem gives us the general term for a binomial expansion. We simplify the powers of x and combine the numerical parts to get a cleaner expression for the general term.

Exam Tip: When simplifying terms like \( (x^2)^{6-r} \), multiply the exponents correctly. Also, remember to handle the \( (-1)^r \) factor if the second term in the binomial is negative.

Question 4. Binomial expression is \( (x^2 – yx)^{12} \).
Answer: For the binomial expression \( (x^2 – yx)^{12} \), we need to find its general term. The general term \( T_{r+1} \) is given by:
\( T_{r+1} = \binom{12}{r} (x^2)^{12-r} (-yx)^r \)
\( = \binom{12}{r} x^{2(12-r)} (-1)^r y^r x^r \)
\( = \binom{12}{r} x^{24-2r} (-1)^r y^r x^r \)
Combining the powers of x, we get:
\( = (-1)^r \binom{12}{r} x^{24-2r+r} y^r \)
\( = (-1)^r \binom{12}{r} x^{24-r} y^r \)
This is the simplified general term for the given binomial expression.
In simple words: We start with the basic formula for the general term, then carefully apply the exponents to each part of the expression. We combine similar terms, especially the powers of x, to get the final general term.

Exam Tip: Always remember to distribute exponents to all parts of a product (e.g., \( (-yx)^r = (-1)^r y^r x^r \)) and combine like bases by adding their exponents.

Question 5. Find the 4th term in the expansion of \( (x – 2y)^{12} \).
Answer: To find the 4th term, we use the formula \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
For the 4th term, \( r+1 = 4 \implies r = 3 \).
Given expression: \( (x – 2y)^{12} \). So, \( n = 12 \), \( a = x \), and \( b = -2y \).
Substituting these values:
\( T_4 = T_{3+1} = \binom{12}{3} (x)^{12-3} (-2y)^3 \)
\( = \binom{12}{3} x^9 (-2)^3 y^3 \)
First, calculate \( \binom{12}{3} \):
\( \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 \)
Next, calculate \( (-2)^3 \):
\( (-2)^3 = -8 \)
Now, put it all together:
\( T_4 = 220 \times x^9 \times (-8) \times y^3 \)
\( T_4 = -1760 x^9 y^3 \)
The 4th term in the expansion is \( -1760 x^9 y^3 \).
In simple words: To find a specific term like the 4th term, we first find the correct 'r' value. Then we use the binomial term formula, putting in 'n', 'a', 'b', and 'r' to calculate the full term, including its sign and powers.

Exam Tip: Remember that for the \( k^{th} \) term, the value of 'r' is \( k-1 \). Also, be careful with negative signs, as an odd power of a negative number will result in a negative outcome.

Question 6. Find the 13th term in the expansion of \( (9x - \frac{1}{3 \sqrt{x}})^{18} \)
Answer: To find the 13th term, we use the formula \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
For the 13th term, \( r+1 = 13 \implies r = 12 \).
Given expression: \( (9x - \frac{1}{3 \sqrt{x}})^{18} \). So, \( n = 18 \), \( a = 9x \), and \( b = -\frac{1}{3 \sqrt{x}} \).
Substitute these values:
\( T_{13} = T_{12+1} = \binom{18}{12} (9x)^{18-12} \left(-\frac{1}{3 \sqrt{x}}\right)^{12} \)
\( = \binom{18}{12} (9x)^6 \left(-\frac{1}{3 x^{1/2}}\right)^{12} \)
\( = \binom{18}{12} (9^6 x^6) \left(\frac{1^{12}}{3^{12} (x^{1/2})^{12}}\right) \)
\( = \binom{18}{12} (9^6 x^6) \left(\frac{1}{3^{12} x^6}\right) \)
Now, we know \( 9 = 3^2 \), so \( 9^6 = (3^2)^6 = 3^{12} \).
\( = \binom{18}{12} (3^{12} x^6) \left(\frac{1}{3^{12} x^6}\right) \)
The \( 3^{12} \) and \( x^6 \) terms cancel out:
\( = \binom{18}{12} \times 1 \)
Now, calculate \( \binom{18}{12} \):
\( \binom{18}{12} = \binom{18}{18-12} = \binom{18}{6} \)
\( = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( = 3 \times 17 \times 2 \times 7 \times 13 \)
\( = 18564 \)
Thus, the 13th term is 18564.
In simple words: We find 'r' for the 13th term and plug all the values into the binomial term formula. Then, we simplify the terms, especially the powers and fractions, making sure to cancel out parts that are the same. Finally, we calculate the combination to get the answer.

Exam Tip: For terms involving \( \sqrt{x} \), convert them to fractional exponents like \( x^{1/2} \) to make simplification easier. Also, remember the property \( \binom{n}{r} = \binom{n}{n-r} \) to simplify calculations for large 'r' values.

Question 7. Number of terms in the expansion is \( 7 + 1 = 8 \). There are two middle terms which are T₄ and T₅. Hence, we are to find T₄ and T₅ in the given expansion \( (3 -\frac{x^{3}}{6})^7 \).
Answer: The given expansion is \( (3 -\frac{x^{3}}{6})^7 \). We can write this as \( [3 + (-\frac{x^{3}}{6})]^7 \).
The general term \( T_{r+1} \) is \( \binom{7}{r} (3)^{7-r} (-\frac{x^{3}}{6})^r \).
To find \( T_4 \), we set \( r+1 = 4 \implies r = 3 \).
\( T_4 = \binom{7}{3} (3)^{7-3} (-\frac{x^{3}}{6})^3 \)
\( = \binom{7}{3} (3)^4 (-\frac{x^{3}}{6})^3 \)
\( = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 3^4 \times (-\frac{1}{6^3}) \times (x^3)^3 \)
\( = 35 \times 81 \times (-\frac{1}{216}) \times x^9 \)
\( = -\frac{35 \times 81}{216} x^9 \)
\( = -\frac{2835}{216} x^9 \)
\( = -\frac{105}{8} x^9 \)
Now, to find \( T_5 \), we set \( r+1 = 5 \implies r = 4 \).
\( T_5 = \binom{7}{4} (3)^{7-4} (-\frac{x^{3}}{6})^4 \)
\( = \binom{7}{4} (3)^3 (-\frac{x^{3}}{6})^4 \)
\( = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} \times 3^3 \times (\frac{1}{6^4}) \times (x^3)^4 \)
\( = 35 \times 27 \times (\frac{1}{1296}) \times x^{12} \)
\( = \frac{35 \times 27}{1296} x^{12} \)
\( = \frac{945}{1296} x^{12} \)
\( = \frac{35}{48} x^{12} \)
The two middle terms are \( T_4 = -\frac{105}{8} x^9 \) and \( T_5 = \frac{35}{48} x^{12} \).
In simple words: When there are an even number of terms in an expansion, there will be two middle terms. We calculate each middle term separately by finding the correct 'r' value for each, then use the general term formula, simplifying all numerical and algebraic parts.

Exam Tip: For an expansion \( (a+b)^n \), if \( n \) is odd, there are \( n+1 \) (an even number of) terms, and the two middle terms are \( T_{(n+1)/2 + 1} \) and \( T_{(n+1)/2} \). Make sure to simplify fractions thoroughly.

Question 8. Find the middle term in the expansion of \( (\frac{x}{3} + 9y)^{10} \).
Answer: The number of terms in the expansion of \( (\frac{x}{3} + 9y)^{10} \) is \( 10 + 1 = 11 \).
Since there is an odd number of terms, there will be only one middle term.
The position of the middle term is \( \frac{10+1}{2} = \frac{11}{2} = 5.5 \), so it's the \( 6^{th} \) term, \( T_6 \).
To find \( T_6 \), we use the formula \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \).
For \( T_6 \), \( r+1 = 6 \implies r = 5 \).
Given expression: \( (\frac{x}{3} + 9y)^{10} \). So, \( n = 10 \), \( a = \frac{x}{3} \), and \( b = 9y \).
Substitute these values:
\( T_6 = T_{5+1} = \binom{10}{5} (\frac{x}{3})^{10-5} (9y)^5 \)
\( = \binom{10}{5} (\frac{x}{3})^5 (9y)^5 \)
\( = \binom{10}{5} \frac{x^5}{3^5} 9^5 y^5 \)
Since \( 9 = 3^2 \), we have \( 9^5 = (3^2)^5 = 3^{10} \).
\( = \binom{10}{5} \frac{x^5}{3^5} 3^{10} y^5 \)
\( = \binom{10}{5} 3^{10-5} x^5 y^5 \)
\( = \binom{10}{5} 3^5 x^5 y^5 \)
Now, calculate \( \binom{10}{5} \):
\( \binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 2 \times 3 \times 2 \times 7 \times 3 = 252 \)
And \( 3^5 = 243 \).
So, \( T_6 = 252 \times 243 x^5 y^5 \)
\( T_6 = 61236 x^5 y^5 \)
The middle term in the expansion is \( 61236 x^5 y^5 \).
In simple words: When the exponent 'n' is even, there is one middle term. We find its position, determine 'r', and then plug all values into the binomial term formula. Simplify powers, especially for terms with coefficients that are multiples of each other, and then calculate the combination.

Exam Tip: For an expansion \( (a+b)^n \), if \( n \) is an even number, there are \( n+1 \) (an odd number of) terms, and the single middle term is \( T_{n/2 + 1} \). Always simplify numerical coefficients like \( 9^5 \) and \( 3^5 \) to their prime factors for easier cancellation.

Question 9. In the expansion of \( (1 + a)^{m+n} \), prove that coefficients of \( a^m \) and \( a^n \) are equal?
Answer: The general term in the expansion of \( (1 + a)^{m+n} \) is given by \( T_{r+1} = \binom{m+n}{r} (1)^{m+n-r} a^r = \binom{m+n}{r} a^r \).
To find the coefficient of \( a^m \), we set \( r = m \):
Coefficient of \( a^m = \binom{m+n}{m} \) (Equation 1)
To find the coefficient of \( a^n \), we set \( r = n \):
Coefficient of \( a^n = \binom{m+n}{n} \) (Equation 2)
We know the identity for binomial coefficients: \( \binom{N}{k} = \binom{N}{N-k} \).
Using this identity for \( \binom{m+n}{m} \), we have:
\( \binom{m+n}{m} = \binom{m+n}{(m+n)-m} = \binom{m+n}{n} \)
Since \( \binom{m+n}{m} \) is equal to \( \binom{m+n}{n} \), it proves that the coefficient of \( a^m \) is equal to the coefficient of \( a^n \).
In simple words: We find the general term of the expansion. Then, we write down the coefficients for \( a^m \) and \( a^n \). Using a rule for combinations, which states that \( \binom{N}{k} \) is the same as \( \binom{N}{N-k} \), we show that these two coefficients are indeed equal.

Exam Tip: This proof relies on the symmetry property of binomial coefficients. Always cite or understand the property \( \binom{n}{r} = \binom{n}{n-r} \) when proving equalities between coefficients.

Question 10. The coefficients of the \( (r – 1)^{th} \), \( r^{th} \) and \( (r + 1)^{th} \) terms in the expansion of \( (x + 1)^n \) are in the ratio of \( 1:3:5 \). Find both n and r.
Answer: The general term in the expansion of \( (x + 1)^n \) is \( T_{k+1} = \binom{n}{k} x^{n-k} (1)^k = \binom{n}{k} x^{n-k} \).
The coefficient of the \( (r-1)^{th} \) term: For \( T_{k+1} = T_{r-1} \), we have \( k+1 = r-1 \implies k = r-2 \).
So, the coefficient is \( \binom{n}{r-2} \). (Equation 1)
The coefficient of the \( r^{th} \) term: For \( T_{k+1} = T_r \), we have \( k+1 = r \implies k = r-1 \).
So, the coefficient is \( \binom{n}{r-1} \). (Equation 2)
The coefficient of the \( (r+1)^{th} \) term: For \( T_{k+1} = T_{r+1} \), we have \( k+1 = r+1 \implies k = r \).
So, the coefficient is \( \binom{n}{r} \). (Equation 3)
According to the problem, the ratio of these coefficients is \( 1:3:5 \):
\( \binom{n}{r-2} : \binom{n}{r-1} : \binom{n}{r} = 1:3:5 \)
From the first two terms:
\( \frac{\binom{n}{r-2}}{\binom{n}{r-1}} = \frac{1}{3} \)
\( \frac{n! / ((r-2)!(n-r+2)!)}{n! / ((r-1)!(n-r+1)!)} = \frac{1}{3} \)
\( \frac{(r-1)!(n-r+1)!}{(r-2)!(n-r+2)!} = \frac{1}{3} \)
\( \frac{(r-1)(r-2)!(n-r+1)!}{(r-2)!(n-r+2)(n-r+1)!} = \frac{1}{3} \)
\( \frac{r-1}{n-r+2} = \frac{1}{3} \)
\( 3(r-1) = n-r+2 \)
\( 3r-3 = n-r+2 \)
\( 4r - 5 = n \) (Equation 4)

From the second and third terms:
\( \frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{3}{5} \)
\( \frac{n! / ((r-1)!(n-r+1)!)}{n! / (r!(n-r)!)} = \frac{3}{5} \)
\( \frac{r!(n-r)!}{(r-1)!(n-r+1)!} = \frac{3}{5} \)
\( \frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!} = \frac{3}{5} \)
\( \frac{r}{n-r+1} = \frac{3}{5} \)
\( 5r = 3(n-r+1) \)
\( 5r = 3n - 3r + 3 \)
\( 8r - 3 = 3n \) (Equation 5)

Now we have a system of two linear equations:
1) \( n = 4r - 5 \)
2) \( 3n = 8r - 3 \)
Substitute n from (Equation 4) into (Equation 5):
\( 3(4r - 5) = 8r - 3 \)
\( 12r - 15 = 8r - 3 \)
\( 12r - 8r = 15 - 3 \)
\( 4r = 12 \)
\( r = 3 \)
Now, substitute \( r = 3 \) back into (Equation 4) to find n:
\( n = 4(3) - 5 \)
\( n = 12 - 5 \)
\( n = 7 \)
Thus, \( n = 7 \) and \( r = 3 \).
In simple words: We list the coefficients of the three consecutive terms using the binomial coefficient formula. Then, we set up two ratios based on the given ratio of \( 1:3:5 \). We solve these two equations together to find the values of 'n' and 'r'.

Exam Tip: When dealing with ratios of binomial coefficients, use the formula \( \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k} \) to simplify the calculations. This approach can save significant time and reduce algebraic errors.

Question 11. Prove that the coefficient of \( x^n \) in \( (1 + x)^{2n} \) is twice the coefficient of \( x^n \) in \( (1 + x)^{2n-1} \).
Answer: First, consider the expansion of \( (1 + x)^{2n} \).
The general term is \( T_{r+1} = \binom{2n}{r} (1)^{2n-r} x^r = \binom{2n}{r} x^r \).
To find the coefficient of \( x^n \), we set \( r = n \).
Coefficient of \( x^n \) in \( (1 + x)^{2n} \) is \( \binom{2n}{n} \). (Equation 1)

Next, consider the expansion of \( (1 + x)^{2n-1} \).
The general term is \( T_{r+1} = \binom{2n-1}{r} (1)^{2n-1-r} x^r = \binom{2n-1}{r} x^r \).
To find the coefficient of \( x^n \), we set \( r = n \).
Coefficient of \( x^n \) in \( (1 + x)^{2n-1} \) is \( \binom{2n-1}{n} \). (Equation 2)

We need to prove that \( \binom{2n}{n} = 2 \times \binom{2n-1}{n} \).
Let's start with the left-hand side (LHS):
\( \binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!} \)

Now, consider the right-hand side (RHS):
\( 2 \times \binom{2n-1}{n} = 2 \times \frac{(2n-1)!}{n!((2n-1)-n)!} \)
\( = 2 \times \frac{(2n-1)!}{n!(n-1)!} \)
To make the denominator of RHS match the LHS, we multiply the numerator and denominator by 'n':
\( = 2 \times \frac{(2n-1)! \times n}{n! (n-1)! \times n} \)
We know that \( n \times (n-1)! = n! \). Also, \( 2n \times (2n-1)! = (2n)! \).
So, we can rewrite the expression as:
\( = \frac{2 \times n \times (2n-1)!}{n! n!} \)
\( = \frac{2n \times (2n-1)!}{n! n!} \)
\( = \frac{(2n)!}{n! n!} \)
Since LHS \( = \frac{(2n)!}{n!n!} \) and RHS \( = \frac{(2n)!}{n!n!} \), we have shown that LHS = RHS.
Therefore, the coefficient of \( x^n \) in \( (1 + x)^{2n} \) is twice the coefficient of \( x^n \) in \( (1 + x)^{2n-1} \). (Proved)
In simple words: We find the coefficient of \( x^n \) for both expressions using the combination formula. Then, we manipulate the expression for the second coefficient to show that it becomes half of the first coefficient, thereby proving the given relationship.

Exam Tip: This proof involves factorial notation. Remember that \( n! = n \times (n-1)! \). A common strategy is to work with the more complex side (usually the one with '2 times' or a fraction) and simplify it until it matches the simpler side.

Question 12. Find a positive value of m for which the coefficient of \( x^2 \) in the expansion of \( (1 + x)^m \) is 6.
Answer: The general term in the expansion of \( (1 + x)^m \) is \( T_{r+1} = \binom{m}{r} (1)^{m-r} x^r = \binom{m}{r} x^r \).
To find the coefficient of \( x^2 \), we set \( r = 2 \).
The coefficient of \( x^2 \) in the expansion of \( (1 + x)^m \) is \( \binom{m}{2} \).
According to the problem, this coefficient is equal to 6:
\( \binom{m}{2} = 6 \)
\( \frac{m(m-1)}{2!} = 6 \)
\( \frac{m(m-1)}{2} = 6 \)
Multiply both sides by 2:
\( m(m-1) = 12 \)
\( m^2 - m = 12 \)
Rearrange into a quadratic equation:
\( m^2 - m - 12 = 0 \)
Factor the quadratic equation:
We need two numbers that multiply to -12 and add to -1. These are -4 and 3.
\( m^2 - 4m + 3m - 12 = 0 \)
\( m(m-4) + 3(m-4) = 0 \)
\( (m-4)(m+3) = 0 \)
This gives two possible values for m:
\( m - 4 = 0 \implies m = 4 \)
\( m + 3 = 0 \implies m = -3 \)
The question asks for a positive value of m. Therefore, \( m = 4 \).
In simple words: We use the binomial coefficient formula to find the expression for the coefficient of \( x^2 \). We set this expression equal to 6, as given. Then, we solve the resulting quadratic equation to find the possible values for 'm' and select the positive one.

Exam Tip: When solving for 'n' or 'm' in binomial problems, you will often get a quadratic equation. Remember to factor or use the quadratic formula to find both possible solutions, and then apply any given conditions (like "positive value") to select the correct answer.