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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Sequences and Series

Let a_ 1 =1 and for n ≥ 1, a_ n+1 =(1)/(2) a_ n + n^ 2 -2 n-1 n^ 2...

Let a1=1a_{1}=1 and for n⩾1,an+1=12an+n2−2n−1n2(n+1)2n \geqslant 1, a_{n+1}=\frac{1}{2} a_{n}+\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}. Then \left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right| is equal to ____\_\_\_\_.

Correct answer

0

Step-by-step solution

Let bn=an−2n2b_n = a_n - \frac{2}{n^2}.
bn+1=12an+n2−2n−1n2(n+1)2−2(n+1)2b_{n+1} = \frac{1}{2}a_n + \frac{n^2-2n-1}{n^2(n+1)^2} - \frac{2}{(n+1)^2}.
Simplifying: bn+1=12bn+(n+1)2+n2−2n−1−2n2n2(n+1)2=12bnb_{n+1} = \frac{1}{2}b_n + \frac{(n+1)^2 + n^2 - 2n - 1 - 2n^2}{n^2(n+1)^2} = \frac{1}{2}b_n.
So bnb_n is GP with ratio 12\frac{1}{2}.
b1=1−2=−1b_1 = 1 - 2 = -1, so bn=−(12)n−1b_n = -\left(\frac{1}{2}\right)^{n-1}.
∑n=1∞bn=−11−12=−2\sum_{n=1}^{\infty} b_n = -\frac{1}{1-\frac{1}{2}} = -2.
\left|\sum_{n=1}^{\infty}\left(a_n - \frac{2}{n^2}\right)\right| = 2$.
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