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

Correct answer: C. ^ 100 · ^ 101 C_ 49 - ^ 101 C_ 50 Let 1 + x = r S = 1 · r + 2 · r^2 + 3 · r^3 + + 100r^ 100 ......(1) (AGP) rS = 1 · r^2 + 2 · r^3 + + 99r^ 100 + 100r^ 101 ......(2) (1) - (2) gives S = - (1+x)^ 101 x^2 + (1)/(x^2) + 100(1+x)^ 101 x coefficient x^ 48 in S = - coefficient of x^ 48 in (1+x)^ 101 x^2 + 100 · Coefficient of x^ 48 in (1+x)^ 101 x = 100 · ^ 101 C_ 49 - ^ 101 C_ 50

JEE Main 202622 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Binomial Theorem

The coefficient of x^ 48 in (1+x)+2(1+x)^ 2 +3(1+x)^ 3 + +100(1+x)^...

The coefficient of

x^{48}

(1+x)+2(1+x)^{2}+3(1+x)^{3}+\ldots+100(1+x)^{100}

is equal to

Options

A $100 \cdot{ }^{100} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{48}$
B ${ }^{100} \mathrm{C}_{50}+{ }^{101} \mathrm{C}_{49}$
C ${ }^{100 \cdot{ }^{101} \mathrm{C}_{49}-{ }^{101} \mathrm{C}_{50}}$
D $100 \cdot{ }^{100} \mathrm{C}_{49}-{ }^{100} \mathrm{C}_{50}$

Correct answer

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

Step-by-step solution

Let

1 + x = r

\therefore S = 1 \cdot r + 2 \cdot r^2 + 3 \cdot r^3 + \ldots + 100r^{100}

......(1) (AGP)

rS = 1 \cdot r^2 + 2 \cdot r^3 + \ldots + 99r^{100} + 100r^{101}

......(2)
(1)

-

(2) gives

S = -\dfrac{(1+x)^{101}}{x^2} + \dfrac{1}{x^2} + \dfrac{100(1+x)^{101}}{x}

\therefore

coefficient

x^{48}

in S

= -

coefficient of

x^{48}

\dfrac{(1+x)^{101}}{x^2} + 100 \cdot

Coefficient of

x^{48}

\dfrac{(1+x)^{101}}{x}

= 100 \cdot {}^{101}C_{49} - {}^{101}C_{50}

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