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ISI2015-MMA-24

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Answer:(B)

$\displaystyle\sum _{k=2}^{\infty}\frac{1}{k(k-1)}$

$\displaystyle=\sum _{k=2}^{\infty}\frac{k-(k-1)}{k(k-1)}$

$\displaystyle=\sum _{k=2}^{\infty}\left(\frac{1}{(k-1)}-\frac{1}{k}\right)$

$=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots$

$=1-\frac{1}{\infty}$

$=1$

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