Binomial Theorem — JEE Main · 2021 · 25 Jul 2021 · Shift 1 Question with Solution

b

is very small as compared to the value of

a

, so that the cube and other higher powers of

\frac{b}{a}

can be neglected in the identity

\frac{1}{a - b} + \frac{1}{a - 2 b} + \frac{1}{a - 3 b} + \dots + \frac{1}{a - nb} = α n + β n^{2} + γ n^{3}

Answer: (C)

Solution

Each term expands as a geometric series in

\frac{b}{a}

\frac{1}{a - k b} = \frac{1}{a} \cdot \frac{1}{1 - \frac{k b}{a}} = \frac{1}{a} (1 + \frac{k b}{a} + \frac{k ^{2} b ^{2}}{a ^{2}} + \dots)

Summing for

k = 1, 2, \dots, n

and keeping powers of

\frac{b}{a}

up to the second,

S = \frac{n}{a} + \frac{b}{a ^{2}} k = 1 \sum n k + \frac{b ^{2}}{a ^{3}} k = 1 \sum n k^{2} = \frac{n}{a} + \frac{b}{a ^{2}} \cdot \frac{n ( n + 1 )}{2} + \frac{b ^{2}}{a ^{3}} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

Only the last sum reaches the third power of

n

, and

\frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{2 n ^{3} + 3 n ^{2} + n}{6}

contributes

\frac{n ^{3}}{3}

. Comparing with

α n + β n^{2} + γ n^{3}

gives

γ = \frac{b ^{2}}{3 a ^{3}}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option