JEE Main 2026 — 21 January, Morning Shift — Question 22

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

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

$\mathrm{n}=1 \mathrm{a}_{2}-\frac{1}{2} \mathrm{a}_{1}=\frac{2}{2^{2}}-\frac{1}{1^{2}}$

$2\left[\mathrm{a}_{3}-\frac{1}{2} \mathrm{a}_{2}=\frac{2}{3^{2}}-\frac{1}{2^{2}}\right]$

$2^{2}\left[a_{4}-\frac{1}{2} a_{3}=\frac{2}{4^{2}}-\frac{1}{3^{2}}\right]$ …..

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

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

Adding $\mathrm{a}_{\mathrm{n}+1}=\frac{2}{(\mathrm{n}+1)^{2}}-\frac{1}{2^{\mathrm{n}}} \Rightarrow \mathrm{a}_{\mathrm{n}}=\frac{2}{\mathrm{n}^{2}}-\frac{1}{2^{\mathrm{n}-1}}$

$\Rightarrow\left|\sum_{\mathrm{n}=1}^{\infty}\left(\mathrm{a}_{\mathrm{n}}-\frac{2}{\mathrm{n}^{2}}\right)\right| \Rightarrow\left|\sum_{\mathrm{n}=1}^{\infty}-\frac{1}{2^{\mathrm{n}-1}}\right|=2$