Question

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

• A. \(100\cdot {101 \choose 49} - {101 \choose 50}\)
• B. \(100\cdot {100 \choose 49} - {100 \choose 48}\)
• C. \(100\cdot {100 \choose 49} - {100 \choose 50}\)
• D. \({100 \choose 50} + {101 \choose 49}\)

Hint:

For sums involving \(k(1+x)^k\), rewrite the sum using derivatives of geometric series to efficiently extract coefficients.

Answer 1

The Correct Option is C

We are given the expression \[ \sum_{k=1}^{100} k(1+x)^k. \]
Step 1: Use a known summation identity.
Recall the identity: \[ \sum_{k=0}^{n} k(1+x)^k = (1+x)\frac{d}{dx}\left(\sum_{k=0}^{n}(1+x)^k\right). \] We know that \[ \sum_{k=0}^{n}(1+x)^k = \frac{(1+x)^{n+1}-1}{x}. \] Thus, \[ \sum_{k=1}^{100} k(1+x)^k = (1+x)\frac{d}{dx}\left(\frac{(1+x)^{101}-1}{x}\right). \]
Step 2: Differentiate the expression.
Differentiating, \[ \frac{d}{dx}\left(\frac{(1+x)^{101}-1}{x}\right) = \frac{x\cdot 101(1+x)^{100} - \big((1+x)^{101}-1\big)}{x^2}. \] Multiplying by \( (1+x) \), \[ \sum_{k=1}^{100} k(1+x)^k = \frac{(1+x)\left[101x(1+x)^{100} - (1+x)^{101} + 1\right]}{x^2}. \]
Step 3: Identify terms contributing to \(x^{48}\).
The coefficient of \(x^{48}\) comes from the expansion of powers of \((1+x)^{100}\) and \((1+x)^{101}\). Extracting the coefficient carefully, we get \[ \text{Coefficient of } x^{48} = 100{100 \choose 49} - {100 \choose 50}. \]
Final Answer: \[ \boxed{100{100 \choose 49} - {100 \choose 50}} \]