Algebra – Binomial Theorem – JEE Main 22 Jan 2026 Shift 1

Solution:

Let the given sum be $S$:
$$ S = 1(1+x) + 2(1+x)^2 + 3(1+x)^3 + \dots + 100(1+x)^{100} $$
This is an Arithmetico-Geometric Progression (AGP). Let $r = 1+x$.
$$ S = 1r + 2r^2 + 3r^3 + \dots + 100r^{100} $$

Multiply by $r$:
$$ rS = 1r^2 + 2r^3 + \dots + 99r^{100} + 100r^{101} $$

Subtract the second equation from the first:
$$ S(1-r) = r + r^2 + r^3 + \dots + r^{100} – 100r^{101} $$

The terms $r + r^2 + \dots + r^{100}$ form a Geometric Progression (GP) with first term $r$, common ratio $r$, and 100 terms.
$$ S(1-r) = \frac{r(r^{100}-1)}{r-1} – 100r^{101} $$

Substitute $r = 1+x$, so $1-r = -x$ and $r-1 = x$:
$$ S(-x) = \frac{(1+x)((1+x)^{100}-1)}{x} – 100(1+x)^{101} $$
$$ -xS = \frac{(1+x)^{101} – (1+x)}{x} – 100(1+x)^{101} $$

Multiply by $-1$ and divide by $x$:
$$ xS = 100(1+x)^{101} – \frac{(1+x)^{101} – (1+x)}{x} $$
$$ S = \frac{100(1+x)^{101}}{x} – \frac{(1+x)^{101}}{x^2} + \frac{1+x}{x^2} $$

We need the coefficient of $x^{48}$ in $S$.
This corresponds to finding:
1. Coefficient of $x^{49}$ in $100(1+x)^{101}$ (due to $1/x$).
2. Coefficient of $x^{50}$ in $-(1+x)^{101}$ (due to $1/x^2$).
The third term $\frac{1+x}{x^2}$ contains only negative powers of $x$, so it contributes 0.

Coefficient calculation:
From first term: $100 \cdot \binom{101}{49}$
From second term: $- \binom{101}{50}$

Total Coefficient $= 100 \cdot ^{101}C_{49} – ^{101}C_{50}$

Ans. (4)