Let a 1 = 1 and for n 1 a n + 1 = 1 2 ? a n + n 2 - 2 n - 1 n 2 ( n + 1 ) 2 . Then | n = 1 ( a n - 2 n 2 ) | is equal to ______. [2026]

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Solution

Q.
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 ______. [2026]

Ans.
(2)

$a_{n + 1} - \frac{1}{2} a_{n} = \frac{n^{2} - 2 n - 1}{n^{2} {(n + 1)}^{2}} = \frac{2 n^{2} - {(n + 1)}^{3}}{n^{2} {(n + 1)}^{2}}$

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

For $n = 1$ ,

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

$2 [a_{3} - \frac{1}{2} a_{2} = \frac{2}{3^{2}} - \frac{1}{2^{2}}]$

$2^{2} [a_{4} - \frac{1}{2} a_{3} = \frac{2}{4^{2}} - \frac{1}{3^{2}}]$

$2^{n - 2} [a_{n} - \frac{1}{2} a_{n - 1} = \frac{2}{n^{2}} - \frac{1}{{(n - 1)}^{2}}]$

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

Adding,

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

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

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