Find the sum of the series:-
√1+112+122+√1+122+132.....√1+120072+120082
a)2008-12008
b)2007-12007
c)2007-12008
d)2008-12007
e)2008-12009
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1 Answers
1T_{n}=\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}} \\ T_{n}=\sqrt{\frac{1}{n^2}+\frac{1}{(n+1)^2}-\frac{2}{n(n+1)}+2(\frac{1}{n}-\frac{1}{n+1})+1}\\ T_{n}=\sqrt{(\frac{1}{n}+\frac{1}{(n+1)})^2+2(\frac{1}{n}-\frac{1}{n+1})+1} \\ \\ \boxed{T_n=\frac{1}{n}-\frac{1}{n+1}+1} \\ \sum_{n=1}^{2007}{T_n}=2007+1-\frac{1}{2008} \\ \boxed{Ans.=2008-\frac{1}{2008}}
