• # question_answer If  $S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)\,!},}$ then $S$ = A) $x+{{x}^{-1}}$ B) $x-{{x}^{-1}}$ C) $\frac{1}{2}(x+{{x}^{-1}})$ D) None of these

Solution :

We have $S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)!}=\left( \frac{{{e}^{\log x}}+{{e}^{-\log x}}}{2} \right)}=\frac{x+{{x}^{-1}}}{2}$.

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