Given that `sum_(n=1)^oo 1/n^2=pi^2/6 and sum_(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b` where a in `N ...
1,350 views · Published 12 October 2018 · 3:43 · Indexed 21 September 2026
Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Given that `sum_(n=1)^oo 1/n^2=pi^2/6 and sum_(n=1)^oo 1/(n^2+8n+16)=pi^2/a-b` where a in `N and b in Q` , then a is equal to
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