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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Sequence and Series
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$\sum_{r=1}^{n}\;\frac{r}{(r+1)!}=$

$(a)\;2-\frac{1}{(n+1)!}\qquad(b)\;1-\frac{1}{(n+1)!}\qquad(c)\;1-\frac{1}{n!}\qquad(d)\;None$

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Answer : (b) $\;1-\frac{1}{(n+1)!}$
$a_{r}=\frac{r}{(r+1)!}=\frac{(r+1)-1}{(r+1)!}$
$=\frac{1}{r!}-\frac{1}{(r+1)!}$
$a_{1}=\frac{1}{1!}-\frac{1}{2!}$
$a_{2}=\frac{1}{2!}-\frac{1}{3!}$
$a_{n}=\frac{1}{n!}-\frac{1}{(n+1)!}$
$S=1-\frac{1}{(n+1)!}.$
answered Jan 21, 2014 by yamini.v
 

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