JEE Main 2026 Jan 21 Shift 1, Mathematics Q22: Summation of Series

JEE Main2026Jan 21, Shift 1Mathematics

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 $n = 1 \sum \infty (a_{n} - \frac{2}{n ^{2}})$ is equal to ______.

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Solution

Rewrite the recurrence: $a_{n + 1} - \frac{1}{2} a_{n} = \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Notice $n^{2} - 2 n - 1 = 2 n^{2} - (n + 1)^{2}$ , so the right side equals $\frac{2}{( n + 1 ) ^{2}} - \frac{1}{n ^{2}}$ .

Multiply the equation for index $n$ by $2^{n - 1}$ and telescope. The closed form turns out to be $a_{n} = \frac{2}{n ^{2}} - \frac{1}{2 ^{n - 1}}$ .

So $a_{n} - \frac{2}{n ^{2}} = - \frac{1}{2 ^{n - 1}}$ , and $n = 1 \sum \infty (a_{n} - \frac{2}{n ^{2}}) = - n = 1 \sum \infty \frac{1}{2 ^{n - 1}} = - 2$ .

The absolute value is $2$ .

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JEE Main 2026 Jan 21 Shift 1, Mathematics Q22: Summation of Series

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