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

Mathematics · Binomial Theorem

JEE Main 2026 — 22 January, Morning Shift — Question 15

The coefficient of $x^{48}$ in $(1+x)+2(1+x)^{2}+3(1 +\mathrm{x})^{3}+\ldots .+100(1+\mathrm{x})^{100}$ is equal to :

Option A:
$100 .{ }^{100} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{50}$
Option B:
${ }^{100} \mathrm{C}_{50}+{ }^{101} \mathrm{C}_{49}$
Option C:
$100 .{ }^{100} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{48}$
Option D:
$100 .{ }^{101} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{50}$
Correct

Answer: D

Step-by-step solution

Let $1+\mathrm{x}=\mathrm{r}$ $\begin{gathered} \therefore \mathrm{S}=1 . \mathrm{r}+2 . \mathrm{r}^{2}+3 . \mathrm{r}^{3}+\ldots \ldots+100 \mathrm{r}^{100} \ldots \ldots \end{gathered}$ $\begin{gathered} \mathrm{rS}=1 \cdot \mathrm{r}^{2}+2 \cdot \mathrm{r}^{3}+\ldots \ldots+99 \mathrm{r}^{100}+100 \mathrm{r}^{101} \end{gathered}$ - gives $S=-\frac{(1+x)^{101}}{x^{2}}+\frac{1}{x^{2}}+\frac{100(1+x)^{101}}{x}$

∴ coefficient $\mathrm{x}^{48}$ in S $=-$ coefficient of $x^{48}$ in $\frac{(1+x)^{101}}{x^{2}}+100$ .

Coefficient of $\mathrm{x}^{48}$ in $\frac{(1+\mathrm{x})^{101}}{\mathrm{x}}$

$=100{ }^{101} \mathrm{C}_{49}-{ }^{101} \mathrm{C}_{50}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Series involving sum & product of Binomial Coefficients