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Q)

Sum of series $\;\sum_{r=1}^{n}\;r\;log\;\frac{r+1}{r}$ is :

$\begin{array}{1 1} n!\;log(n+1) \\ log\;\frac{n+1}{n!} \\n\;log(n+1)-log\;n! \\ None \end{array}$

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A)
Answer : (c) $n\;log(n+1)-log\;n!$
Explanation : $a_{r}=r\;log\;\frac{r+1}{r}$
$=r\;[log(r+1)-log\;r]$
$=(r+1)\;log(r+1)-r\;log\;r-log(r+1)$
$S_{n}=\sum_{r=1}^{n}\;(r+1)\;log(r+1)-r\;log\;r-\sum_{r=1}^{n}\;log(r+1)$
$=(n+1)\;log(n+1)-1\;log\;1-log(n+1)!$
$=log\;\frac{(n+1)^{n+1}}{(n+1)!}=log\;\frac{(n+1)^n}{n!}$
$=n\;log(n+1)-log\;n!$
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