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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Sequence and Series

Sum of $\;\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+\;.....\;upto\;\infty\;is$

$(a)\;\frac{1}{24}\qquad(b)\;\frac{1}{18}\qquad(c)\;\frac{1}{27}\qquad(d)\;\frac{1}{6}$

1 Answer

Answer : (b) $\;\frac{1}{18}$
Explanation : $a_{r}=\frac{1}{r(r+1)(r+2)(r+3)}$
$=\frac{1}{3}\;\frac{(r+3)-r}{r(r+1)(r+2)(r+3)}$
$=\frac{1}{3} \;[\frac{1}{r(r+1)(r+2)}-\frac{1}{(r+1)(r+2)(r+3)}]$
$a_{1}=\frac{1}{3}\;[\frac{1}{1.2.3}-\frac{1}{2.3.4}]$
$a_{2}=\frac{1}{3}\;[\frac{1}{2.3.4}-\frac{1}{3.4.5}]$
$S=\frac{1}{3}*\frac{1}{1.2.3}=\frac{1}{18}.$
answered Jan 21, 2014 by yamini.v
 

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