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

To find the coefficient of \(x^{48}\) in the given expression:

\[ (1+x) + 2(1+x)^2 + 3(1+x)^3 + \cdots + 100(1+x)^{100} \]

This expression can be rewritten using summation notation as:

\[ \sum_{k=1}^{100} k(1+x)^k \]

We need to find the coefficient of \(x^{48}\) in the entire summation above. Start by evaluating the expansion of each term \(k(1+x)^k\). The binomial expansion of \((1+x)^k\) is:

\[ (1+x)^k = \sum_{j=0}^{k} \binom{k}{j} x^j \]

Thus, each term contributes:

\[ k \cdot \binom{k}{j} x^j \]

For the desired coefficient of \(x^{48}\), we want:

\[ \sum_{k=48}^{100} k \cdot \binom{k}{48} \]

Because only terms where \(k \geq 48\) contribute to the coefficient of \(x^{48}\) and above.

Thus, we explicitly compute this using the identity that relates the sum of binomial coefficients:

\[ \sum_{k=48}^{100} k \cdot \binom{k-1}{47} \]

The expression simplifies to leveraging summation and manipulation:

\[ \sum_{k=1}^{100} \sum_{j=0}^{k} \binom{k}{j} x^j \]

The focused component is programmed by:

\[ \sum_{k=49}^{100} \binom{k}{48} = \binom{100+1}{49} \]

The sum formula then applies, accounting back shift:

\[ \sum_{k=1}^{100} \binom{k-1}{48} = 100 \cdot \binom{100}{49} - \binom{100}{50} \]

Therefore, the coefficient of \(x^{48}\) is:

\[ 100 \cdot \binom{100}{49} - \binom{100}{50} \]

So, the correct answer is \(\boxed{100 \cdot {100 \choose 49} - {100 \choose 50}}\).