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JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Sequences and Series

Let

a_{1} = 1

and for

n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}

. Then

\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})

is equal to

Answer: 2

Solution

The recurrence rewrites as

a_{n + 1} - \frac{1}{2} a_{n} = \frac{2}{( n + 1 ) ^{2}} - \frac{1}{n ^{2}}

, and

a_{n} = \frac{2}{n ^{2}} - \frac{1}{2 ^{n - 1}}

satisfies both this relation and

a_{1} = 1

. Hence

a_{n} - \frac{2}{n ^{2}} = - \frac{1}{2 ^{n - 1}}

, so

\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}}) = - \sum_{n = 1}^{\infty} \frac{1}{2 ^{n - 1}} = 2

.

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Integer answer

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