JEE Main 2026 Jan 22 Shift 1, Mathematics Q16: Arithmetic, Geometric And Harmonic Progressions

JEE Main2026Jan 22, Shift 1Mathematics

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

Solution

Let $r = 1 + x$ and $S = \sum_{k = 1}^{100} k r^{k}$ . This is an arithmetico-geometric progression; multiply by $r$ and subtract:

$S - r S = (r + r^{2} + \dots + r^{100}) - 100 r^{101}$ .

Solving with $r - 1 = x$ gives $S = - \frac{( 1 + x ) ^{101}}{x ^{2}} + \frac{1}{x ^{2}} + \frac{100 ( 1 + x ) ^{101}}{x}$ .

Extract coefficient of $x^{48}$ :

From $\frac{100 ( 1 + x ) ^{101}}{x}$ : equals 100 $\times$ (coefficient of $x^{49}$ in $(1 + x)^{101}$ ) $= 100 \cdot^{101} C_{49}$ .

From $- \frac{( 1 + x ) ^{101}}{x ^{2}}$ : equals $- 1 \times$ (coefficient of $x^{50}$ in $(1 + x)^{101}$ ) $= -^{101} C_{50}$ .

The $\frac{1}{x ^{2}}$ term has only the $x^{- 2}$ contribution, nothing at $x^{48}$ .

Total: $100 \cdot^{101} C_{49} -^{101} C_{50}$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.