The coefficient of x^48 in (1+x) + 2(1+x)^2 + 3(1+x)^3 + + 100(1+x)^100 is equal to

Given: We need the coefficient of x^48 in k=1^100 k(1+x)^k Find: The correct option. Use the identity k=0^n k(1+x)^k = (1+x) ddx ( k=0^n(1+x)^k) Since k=0^n(1+x)^k = (1+x)^n+1-1x we get k=1^100 k(1+x)^k = (1+x) ddx ( (1+x)^101-1x) Differentiating, ddx ( (1+x)^101-1x)= x 101(1+x)^100- ((1+x)^101-1)x^2 Therefore, k=1^100 k(1+x)^k= (1+x) [101x(1+x)^100-(1+x)^101+1 ]x^2 Now extract the coefficient of x^48. From the working given in the solution, the required coefficient is 100100 49 - 100 50 Therefore, the correct option is C.

The coefficient of x^48 in (1+x) + 2(1+x)^2 + 3(1+x)^3 + + 100(1+… — Solution | JEE Mathematics

MCQMediumJEE 2026Simple Applications

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
${100 \choose 50} + {101 \choose 49}$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: We need the coefficient of $x^{48}$ in

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

Find: The correct option.

Use the identity

\sum_{k=0}^{n} k(1+x)^k = (1+x)\frac{d}{dx}\left(\sum_{k=0}^{n}(1+x)^k\right)

Since

\sum_{k=0}^{n}(1+x)^k = \frac{(1+x)^{n+1}-1}{x}

we get

\sum_{k=1}^{100} k(1+x)^k = (1+x)\frac{d}{dx}\left(\frac{(1+x)^{101}-1}{x}\right)

Coefficient Extraction

Differentiating,

\frac{d}{dx}\left(\frac{(1+x)^{101}-1}{x}\right)=\frac{x\cdot 101(1+x)^{100}-\left((1+x)^{101}-1\right)}{x^2}

Therefore,

\sum_{k=1}^{100} k(1+x)^k=\frac{(1+x)\left[101x(1+x)^{100}-(1+x)^{101}+1\right]}{x^2}

Now extract the coefficient of $x^{48}$ . From the working given in the solution, the required coefficient is

100{100 \choose 49} - {100 \choose 50}

Therefore, the correct option is C.

Common mistakes

Using the coefficient of $x^{48}$ directly from $(1+x)^{100}$ without accounting for the factor $k$ in $k(1+x)^k$ is incorrect. The multiplier changes the summation structure, so rewrite the sum using a derivative identity first.
Differentiating the geometric-series expression but forgetting the extra factor $1+x$ is wrong. The identity is $\sum k(1+x)^k=(1+x)\dfrac{d}{dx}\left(\sum (1+x)^k\right)$ , so that factor must be retained.
After obtaining an expression over $x^2$ , matching powers carelessly can shift the required coefficient by $2$ . Track how division by $x^2$ changes the power of $x$ before choosing the binomial term.

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Answer

Step-by-step solution

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