Date, Time & Arrangement Problems If xy\neq0 and x*y=1+1/{xy}, then \\sum^{2020}_{a=1} (1/a^2*a)=?

Correct Answer is : 2023 x 1010

AP ICET 11 Sep 2020 Shift 2 Paper

Question : 74 of 200

Marks: +1, -0

Date, Time & Arrangement Problems

xy≠0

and

x*y=1+1/{xy}

, then

\Σ↖{2020}↙{a=1} (1/a^2*a)=?

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