The value of S=1+112+122+1+122+132+..+1+1(2014)2+1(2015)2 is
2015
2015−12015
2014−12014
2016−12016
Tn=1+1n2+1(n+1)2=n2+(n+1)2+n2(n+1)2n2(n+1)2=2n2+2n+1+n2(n+1)2n2(n+1)2=n(n+1)+1n(n+1)=1+1n−1n+1S=∑x=12014 1+1n−1n+1=2014+1−12015=2015−12015