If the coefficients of x7 in (ax2+2bx1)11 and x−7 in (ax−3bx21)11 are equal, then
Held on 6 Apr 2023 · Verified 6 Jul 2026.
729ab=32
32ab=729
64ab=243
243ab=64
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Let $a_{1}=1$ and for $n \geqslant 1, a_{n+1}=\frac{1}{2} a_{n}+\frac{n^{2}-2 n-1}{n^{2}(n+1)^{2}}$. Then $\left|\sum_{n=1}^{\infty}\left(a_{n}-\frac{2}{n^{2}}\right)\right|$ is equal to $\_\_\_\_$.
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