RBSE Solutions Class 11 Maths Chapter 7 Binomial Theorem Exercise 7.3

Question 1. If \( C_0, C_1, C_2, \ldots, C_n \) are coefficients of expansion \( (1 + x)^n \) then find the value of:
(i) \( {}^8C_1 + {}^8C_2 + {}^8C_3 + \ldots + {}^8C_8 \)
(ii) \( {}^8C_1 + {}^8C_3 + {}^8C_5 + {}^8C_7 \)
Answer:
(i) To find the sum \( {}^8C_1 + {}^8C_2 + {}^8C_3 + \ldots + {}^8C_8 \):
We know the identity: \( {}^nC_0 + {}^nC_1 + {}^nC_2 + \ldots + {}^nC_n = 2^n \).
Here, \( n = 8 \).
So, \( {}^8C_0 + {}^8C_1 + {}^8C_2 + \ldots + {}^8C_8 = 2^8 \).
We know that \( {}^8C_0 = 1 \).
\( \implies 1 + ( {}^8C_1 + {}^8C_2 + {}^8C_3 + \ldots + {}^8C_8 ) = 2^8 \)
\( \implies {}^8C_1 + {}^8C_2 + {}^8C_3 + \ldots + {}^8C_8 = 2^8 - 1 \)
\( \implies {}^8C_1 + {}^8C_2 + {}^8C_3 + \ldots + {}^8C_8 = 256 - 1 = 255 \).
(ii) To find the sum \( {}^8C_1 + {}^8C_3 + {}^8C_5 + {}^8C_7 \):
We know the identity for the sum of odd binomial coefficients: \( {}^nC_1 + {}^nC_3 + {}^nC_5 + \ldots = 2^{n-1} \).
Here, \( n = 8 \).
So, \( {}^8C_1 + {}^8C_3 + {}^8C_5 + {}^8C_7 = 2^{8-1} \)
\( \implies {}^8C_1 + {}^8C_3 + {}^8C_5 + {}^8C_7 = 2^7 \)
\( \implies {}^8C_1 + {}^8C_3 + {}^8C_5 + {}^8C_7 = 128 \).
In simple words: For part (i), we use a rule that says all combinations for a power of 'n' add up to \( 2^n \). Since the first term \( {}^nC_0 \) is 1, we subtract 1 from \( 2^n \). For part (ii), there's a special rule for only the odd-numbered combinations (like \( C_1, C_3 \), etc.), which states their sum is \( 2^{n-1} \).

🎯 Exam Tip: Remember the basic binomial identities for sums of coefficients. \( \sum_{r=0}^n {}^nC_r = 2^n \) and \( \sum_{r \text{ odd}} {}^nC_r = \sum_{r \text{ even}, r \neq 0} {}^nC_r = 2^{n-1} \).

Question 2. Prove that \( C_0 + 3C_1 + 5C_2 + \ldots + (2n + 1)C_n = (n + 1)2^n \).
Answer:
Let's start with the Left Hand Side (L.H.S.):
L.H.S. \( = C_0 + 3C_1 + 5C_2 + \ldots + (2n + 1)C_n \)
We can write each term \( (2r + 1)C_r \) as \( (2r + 1){}C_r \).
L.H.S. \( = \sum_{r=0}^n (2r + 1)C_r \)
This can be split into two sums:
\( = \sum_{r=0}^n 2r C_r + \sum_{r=0}^n 1 C_r \)
We know the identity: \( rC_r = nC_{r-1} \). (This means \( \sum_{r=1}^n rC_r = n2^{n-1} \))
The first sum becomes: \( 2 \sum_{r=1}^n rC_r \) (because for \( r=0 \), \( rC_r = 0 \))
\( = 2 (n2^{n-1}) + \sum_{r=0}^n C_r \)
The second sum is the sum of all binomial coefficients: \( \sum_{r=0}^n C_r = 2^n \).
\( = 2n2^{n-1} + 2^n \)
\( = n2^n + 2^n \)
Now, we can take \( 2^n \) as common:
\( = (n + 1)2^n \).
This is equal to the Right Hand Side (R.H.S.).
Hence Proved.
In simple words: To prove this, we separate the sum into two parts. One part uses a special rule that \( rC_r \) sums up to \( n2^{n-1} \). The other part is just the sum of all \( C_r \) terms, which is \( 2^n \). Adding these parts together and simplifying gives us the answer we needed to prove.

🎯 Exam Tip: When terms like \( (2r+1)C_r \) appear, try splitting them into \( 2rC_r + C_r \). Remember the key identities \( \sum_{r=0}^n C_r = 2^n \) and \( \sum_{r=1}^n rC_r = n2^{n-1} \).

Question 3. Prove that \( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n = \frac{(2n)!}{(n-2)!(n+2)!} \).
Answer:
From the Binomial Theorem, we know:
\( (1 + x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_rx^r + \ldots + C_nx^n \) ... (i)
Now, consider the expansion of \( (x + 1)^n \) in reverse order:
\( (x + 1)^n = C_0x^n + C_1x^{n-1} + C_2x^{n-2} + \ldots + C_rx^{n-r} + \ldots + C_n \) ... (ii)
Multiply equation (i) and (ii):
\( (1 + x)^{2n} = (C_0 + C_1x + C_2x^2 + \ldots + C_nx^n) \times (C_0x^n + C_1x^{n-1} + C_2x^{n-2} + \ldots + C_n) \)
We are looking for the coefficient of \( x^{n-2} \) in the product. The terms that contribute to \( x^{n-2} \) are formed by multiplying a term \( C_k x^k \) from the first expansion with a term \( C_j x^{n-j} \) from the second expansion such that \( k + n - j = n - 2 \). This simplifies to \( k - j = -2 \), or \( j = k + 2 \).
So, the coefficient of \( x^{n-2} \) will be:
\( C_0C_2 + C_1C_3 + C_2C_4 + \ldots + C_{n-2}C_n \).
From the expansion of \( (1 + x)^{2n} \), the coefficient of \( x^{n-2} \) is \( {}^{2n}C_{n-2} \).
Therefore, we have:
\( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n = {}^{2n}C_{n-2} \).
Now, let's write \( {}^{2n}C_{n-2} \) using the factorial formula \( {}^nC_r = \frac{n!}{r!(n-r)!} \):
\( {}^{2n}C_{n-2} = \frac{(2n)!}{(n-2)! (2n - (n-2))!} \)
\( \implies {}^{2n}C_{n-2} = \frac{(2n)!}{(n-2)! (2n - n + 2)!} \)
\( \implies {}^{2n}C_{n-2} = \frac{(2n)!}{(n-2)! (n+2)!} \).
Thus, \( C_0 C_2 + C_1 C_3 + C_2 C_4 + \ldots + C_{n-2} C_n = \frac{(2n)!}{(n-2)!(n+2)!} \).
Hence Proved.
In simple words: We multiply two binomial expansions, \( (1+x)^n \) and \( (x+1)^n \), to get \( (1+x)^{2n} \). We then find the part that has \( x^{n-2} \) in this combined expansion. By comparing the coefficients, we find that the sum of products matches the binomial coefficient \( {}^{2n}C_{n-2} \). Finally, we write this coefficient using factorials, which gives us the required result.

🎯 Exam Tip: For sums of products of binomial coefficients, \( \sum C_i C_j \), often consider the coefficient of \( x^k \) in \( (1+x)^n (x+1)^n \). The form \( (x+1)^n \) helps generate the required terms for matching powers.

Question 5. Prove that \( \left(1 + \frac{C_1}{C_0}\right) \left(1 + \frac{C_2}{C_1}\right) \left(1 + \frac{C_3}{C_2}\right) \ldots \left(1 + \frac{C_n}{C_{n-1}}\right) = \frac{(n+1)^n}{n!} \).
Answer:
Let's consider a general term in the product: \( 1 + \frac{C_r}{C_{r-1}} \).
We know the identity \( \frac{C_r}{C_{r-1}} = \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n!}{r!(n-r)!} \times \frac{(r-1)!(n-r+1)!}{n!} = \frac{n-r+1}{r} \).
So, substitute this into the general term:
\( 1 + \frac{C_r}{C_{r-1}} = 1 + \frac{n-r+1}{r} \)
\( = \frac{r + (n-r+1)}{r} \)
\( = \frac{n+1}{r} \).
Now, let's substitute this back into the original product:
For \( r=1 \): \( 1 + \frac{C_1}{C_0} = \frac{n+1}{1} \)
For \( r=2 \): \( 1 + \frac{C_2}{C_1} = \frac{n+1}{2} \)
For \( r=3 \): \( 1 + \frac{C_3}{C_2} = \frac{n+1}{3} \)
...
For \( r=n \): \( 1 + \frac{C_n}{C_{n-1}} = \frac{n+1}{n} \)
Now, multiply all these terms together:
L.H.S. \( = \left(\frac{n+1}{1}\right) \left(\frac{n+1}{2}\right) \left(\frac{n+1}{3}\right) \ldots \left(\frac{n+1}{n}\right) \)
\( = \frac{(n+1) \times (n+1) \times (n+1) \times \ldots \times (n+1)}{1 \times 2 \times 3 \times \ldots \times n} \)
Since there are \( n \) terms in the product, \( (n+1) \) is multiplied \( n \) times in the numerator.
\( = \frac{(n+1)^n}{n!} \).
This is equal to the Right Hand Side (R.H.S.).
Hence Proved.
In simple words: We look at each part of the multiplication separately. Each part, like \( (1 + C_r/C_{r-1}) \), simplifies to \( (n+1)/r \) using a known binomial coefficient identity. When we multiply all these simplified parts together, we get \( (n+1) \) multiplied by itself 'n' times in the top, and \( 1 \times 2 \times 3 \times \ldots \times n \) in the bottom, which is \( n! \). This gives us the final answer.

🎯 Exam Tip: When you see ratios of binomial coefficients, immediately think of the identity \( \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r} \). This is a powerful tool for simplifying such expressions.

Question 6. If expansion of \( (1 + x - 2x^2)^6 \) is denoted by \( 1 + a_1x + a_2x^2 + a_3x^3 + \ldots + a_{12}x^{12} \) then prove that \( a_2 + a_4 + a_6 + \ldots + a_{12} = 31 \).
Answer:
Let the given expansion be:
\( (1 + x - 2x^2)^6 = 1 + a_1x + a_2x^2 + a_3x^3 + \ldots + a_{12}x^{12} \) ... (i)
First, substitute \( x = 1 \) into equation (i):
\( (1 + 1 - 2(1)^2)^6 = 1 + a_1(1) + a_2(1)^2 + a_3(1)^3 + \ldots + a_{12}(1)^{12} \)
\( \implies (2 - 2)^6 = 1 + a_1 + a_2 + a_3 + \ldots + a_{12} \)
\( \implies 0^6 = 1 + a_1 + a_2 + a_3 + \ldots + a_{12} \)
\( \implies 1 + a_1 + a_2 + a_3 + \ldots + a_{12} = 0 \) ... (ii)
Next, substitute \( x = -1 \) into equation (i):
\( (1 + (-1) - 2(-1)^2)^6 = 1 + a_1(-1) + a_2(-1)^2 + a_3(-1)^3 + \ldots + a_{12}(-1)^{12} \)
\( \implies (1 - 1 - 2(1))^6 = 1 - a_1 + a_2 - a_3 + \ldots + a_{12} \)
\( \implies (-2)^6 = 1 - a_1 + a_2 - a_3 + \ldots + a_{12} \)
\( \implies 64 = 1 - a_1 + a_2 - a_3 + \ldots + a_{12} \) ... (iii)
Now, add equation (ii) and equation (iii):
\( (1 + a_1 + a_2 + a_3 + \ldots + a_{12}) + (1 - a_1 + a_2 - a_3 + \ldots + a_{12}) = 0 + 64 \)
Notice that all odd-powered terms (\( a_1, a_3 \), etc.) will cancel out.
\( \implies 2(1 + a_2 + a_4 + \ldots + a_{12}) = 64 \)
Divide both sides by 2:
\( \implies 1 + a_2 + a_4 + \ldots + a_{12} = 32 \)
Finally, subtract 1 from both sides:
\( \implies a_2 + a_4 + a_6 + \ldots + a_{12} = 32 - 1 \)
\( \implies a_2 + a_4 + a_6 + \ldots + a_{12} = 31 \).
This proves the required result.
In simple words: To find the sum of even-indexed coefficients, we use a clever trick. First, we substitute \( x=1 \) into the given expansion to get an equation with all coefficients. Then, we substitute \( x=-1 \) to get another equation where odd-indexed coefficients become negative. When we add these two equations, all the odd-indexed terms cancel out, leaving us with twice the sum of the even-indexed terms plus twice the constant term. From this, we can easily find the sum we need.

🎯 Exam Tip: For problems asking for sums of even or odd coefficients, always consider substituting \( x=1 \) and \( x=-1 \) into the original expansion and then adding or subtracting the resulting equations. This is a standard method to isolate desired coefficients.