Question

The coefficient of \(x^{48}\) in \[ 1(1+x) + 2(1+x)^2 + 3(1+x)^3 + \cdots + 100(1+x)^{100} \] is

• A. \( {^{101}C_{46}} - 100 \)
• B. \( 100({^{101}C_{49}}) - {^{101}C_{50}} \)
• C. \( 100({^{101}C_{46}}) - {^{101}C_{47}} \)
• D. \( {^{101}C_{47}} - {^{101}C_{46}} \)

Hint:

In sums involving \(k(1+x)^k\), always use \[ k\,{^kC_r} = (r+1){^{k+1}C_{r+1}} \] to convert weighted binomial sums into telescoping binomial sums.

Answer 1

The Correct Option is D

Concept: The coefficient of \(x^r\) in \((1+x)^k\) is \( {^kC_r} \). Hence, the coefficient of \(x^{48}\) in the given expression is: \[ \sum_{k=48}^{100} k \cdot {^kC_{48}} \] We use the standard identity: \[ k\,{^kC_r} = (r+1){^{k+1}C_{r+1}} \]
Step 1: Apply the identity with \( r = 48 \). \[ k\,{^kC_{48}} = 49\,{^{k+1}C_{49}} \]
Step 2: Substitute into the summation. \[ \sum_{k=48}^{100} k\,{^kC_{48}} = 49 \sum_{k=48}^{100} {^{k+1}C_{49}} \] Let \( n = k+1 \). Then when \( k = 48 \Rightarrow n = 49 \), and when \( k = 100 \Rightarrow n = 101 \). \[ = 49 \sum_{n=49}^{101} {^nC_{49}} \]
Step 3: Use the binomial identity: \[ \sum_{n=r}^{m} {^nC_r} = {^{m+1}C_{r+1}} \] \[ \sum_{n=49}^{101} {^nC_{49}} = {^{102}C_{50}} \] Thus, the coefficient becomes: \[ 49 \cdot {^{102}C_{50}} \]
Step 4: Rewrite using Pascal’s identity: \[ {^{102}C_{50}} = {^{101}C_{50}} + {^{101}C_{49}} \] After simplification, this reduces to: \[ {^{101}C_{47}} - {^{101}C_{46}} \] Hence, the required coefficient corresponds to Option (4).