The coefficient of x^{48} in (1+x)+2(1+x)²+3(1+x)³++100(1+x)¹00} is equal to [22 Jan 2026 Shift 1]

Correct Answer is : 100⋅(10149)−(10150)100 · {101}{49} - {101}{50}100⋅(49101)−(50101)

Correct Answer is : 100⋅(10149)−(10150)100 · {101}{49} - {101}{50}100⋅(49101)−(50101)

Test Index

Binomial Theorem Part 3

Section: Mathematics

Question : 15 of 53

Marks: +1, -0

The coefficient of

x^{48}

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

Solution:

Concept:

The series is an arithmetico‑geometric progression in

(1+x)

. Subtract

(1+x)S

from

S

to find a closed form.

Explanation:

Let

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

Multiply by

(1+x)

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

Subtract:

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

-xS = (1+x)\frac{(1+x)^{100}-1}{x} - 100(1+x)^{101}

Thus

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

Coefficient of

x^{48}

S

comes from:

• First term:

-\frac{(1+x)^{101}}{x^2}

gives

-\binom{101}{50}

• Second term:

\frac{100(1+x)^{101}}{x}

gives

100\binom{101}{49}

Hence required coefficient =

100\binom{101}{49} - \binom{101}{50}

Answer:

Option B.

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