Let us observe that the general n^(th)" term "T_nnth term Tn of the given
Series is given by,
T_n^2=1+1/n^2+1/(n+1)^2T2n=1+1n2+1(n+1)2
={n^2(n+1)^2+(n+1)^2+n^2]/{n^2(n+1)^2}=n2(n+1)2+(n+1)2+n2n2(n+1)2
=[{n(n+1)}^2+n^2+2n+1+n^2]/{n(n+1)}^2={n(n+1)}2+n2+2n+1+n2{n(n+1)}2
=[{n(n+1)}^2+2n^2+2n+1]/{n(n+1)}^2={n(n+1)}2+2n2+2n+1{n(n+1)}2
=[{n(n+1)}^2+2(1){n(n+1)}+1^2]/[{n(n+1)}^2}={n(n+1)}2+2(1){n(n+1)}+12{n(n+1)}2
={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=[n(n+1)+1n(n+1)]2
rArr T_n={n(n+1)+1}/{n(n+1)}={n(n+1))/{n(n+1))+1/{n(n+1)}⇒Tn=n(n+1)+1n(n+1)=n(n+1)n(n+1)+1n(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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