Question

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

• A. \( (101C46) - 100 \)
• B. \( 100(101C49) - 101C50 \)
• C. \( 100(101C46) - 101C47 \)
• D. \( 101C47 - 101C46 \)

Hint:

In expansions of this type, each term contributes to the power of \( x \) based on the binomial theorem.

Answer 1

The Correct Option is B

Step 1: General term in the expansion.
The general term in the expansion is: \[ n(1+x)^n \] where \( n \) varies from 1 to 100. Step 2: Finding the coefficient of \( x^{48} \).
To find the coefficient of \( x^{48} \), use the binomial expansion for each term and sum the contributions for \( x^{48} \) from each term. For the general term \( n(1+x)^n \), the coefficient of \( x^{48} \) is: \[ n \times \binom{n}{48} \] Step 3: Summing the contributions.
The coefficient of \( x^{48} \) in the full expansion is given by the sum of the contributions from all terms: \[ 100 \times \binom{101}{49} - 101 \times \binom{100}{50} \] Step 4: Conclusion.
Thus, the correct coefficient is \( 100(101C49) - 101C50 \). Final Answer: \[ \boxed{100(101C49) - 101C50} \]