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

Given $a_{n}=\sum_{k=1}^{n}\;\sqrt{1+\large\frac{1}{k^2}+\large\frac{1}{(k+1)^2}}\;$ the value of $\;a_{5}\;$ is

$(a)\;35\qquad(b)\;\large\frac{35}{6}\qquad(c)\;36\qquad(d)\;6$

1 Answer

Answer : (b) $\;\large\frac{35}{6}$
Explanation : $\;1+\large\frac{1}{k^2}+\large\frac{1}{(k+1)^2}$
$=\large\frac{k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}$
$=\large\frac{(k^2+k+1)}{k^2+(k+1)^2}$
$a_{n}=\sum_{k=1}^{n}\;\large\frac{k^2+k+1}{k (k+1)}$
$\sum_{k=1}^{n}=\large\frac{k(k+1)+1}{k(k+1)}$
$=\sum_{k=1}^{n}\;1+\large\frac{1}{k}-\large\frac{1}{k+1}$
$a_{n}=n+1-\large\frac{1}{n+1}$
$a_{5}=5+1-\large\frac{1}{5+1}$
$=6-\large\frac{1}{6}=\large\frac{35}{6}.$
answered Jan 23, 2014 by yamini.v
 

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