The value of $$1+112+122+1+122+132+$$..$$+1+1(2020)2+1(2021)2$$ is

Correct option is 22021−12021

Questions

$The value of \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . + \sqrt{1 + \frac{1}{(2020)^{2}} + \frac{1}{(2021)^{2}}} is$

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2021

2021−12021

2020−12020

2021−12020

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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+1required sum=S=∑x=12020 1+1n−1n+1=2020+1−12021=2021−12021

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The value of 1+112+122+1+122+132+..+1+1(2020)2+1(2021)2 is

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Ready to Test Your Skills?

free study material

$The value of \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . + \sqrt{1 + \frac{1}{(2020)^{2}} + \frac{1}{(2021)^{2}}} is$