Question #8d376

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Answer 1 · 2017-02-06T17:32:22

Refer to the Explanation given below for the Answer : $2017-1/2017.$

Let us observe that the general $n^(th)" term "T_n$ of the given

Series is given by,

$T_n^2=1+1/n^2+1/(n+1)^2$

$={n^2(n+1)^2+(n+1)^2+n^2]/{n^2(n+1)^2}$

$=[{n(n+1)}^2+n^2+2n+1+n^2]/{n(n+1)}^2$

$=[{n(n+1)}^2+2n^2+2n+1]/{n(n+1)}^2$

$=[{n(n+1)}^2+2(1){n(n+1)}+1^2]/[{n(n+1)}^2}$

$={n(n+1)+1}^2/[{n(n+1)}^2}$

$=[{n(n+1)+1}/{n(n+1)}]^2$

$rArr T_n={n(n+1)+1}/{n(n+1)}={n(n+1))/{n(n+1))+1/{n(n+1)}$

$rArr T_n=1+{(n+1)-n}/{n(n+1)}=1+cancel(n+1)/{n(cancel(n+1))}-canceln/{canceln(n+1)}$

$:. T_n=1+1/n-1/(n+1)$

Hence, $S_n=sumT_n=sum[1+1/n-1/(n+1)]$

$=sum1+sum{1/n-1/(n+1)}=n+sum{1/n-1/(n+1)}$

$=n+{(1/1cancel(-1/2))+(cancel(1/2)cancel(-1/3))+(cancel(1/3)cancel(-1/4))+...+(cancel(1/(n-1))cancel(-1/n))+(cancel(1/n)-1/(n+1)}$

$"So, in general, "S_n=n+{1-1/(n+1)}=(n+1)-1/((n+1)).$

In particular, $S_2016=2017-1/2017.$

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