The value of $$S=1+112+122+1+122+132+$$..$$+1+1(2014)2+1(2015)2$$ is

Correct option is 22015−12015

Questions

$The value of S = \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . + \sqrt{1 + \frac{1}{(2014)^{2}} + \frac{1}{(2015)^{2}}} is$

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2015−12015

2014−12014

2016−12016

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detailed solution

Correct option is B

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

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The value of S=1+112+122+1+122+132+..+1+1(2014)2+1(2015)2 is

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$The value of S = \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . + \sqrt{1 + \frac{1}{(2014)^{2}} + \frac{1}{(2015)^{2}}} is$