Find the sum of -1-1-12-1-22-1-1-22-1-32- - -1-1-20072-1-20082-

The correct option is A 2008 1/2008Option (A) is the correct answer. Check the video for the approach. ...

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Integration Using Substitution

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Question

Find the sum of $\sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + . . . . . + \sqrt{1 + \frac{1}{2007^{2}} + \frac{1}{2008^{2}}}$ .

2008 - \frac{1}{2008}

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2007 - \frac{1}{2007}

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2007 - \frac{1}{2008}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2008 - \frac{1}{2007}

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\sqrt{(1 + \frac{1}{1^{2}} + \frac{1}{2^{2}})} + \sqrt{(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}})} + . . . . . . . . . + \sqrt{(1 + \frac{1}{2007^{2}} + \frac{1}{2008^{2}})}

\sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + \sqrt{1 + \frac{1}{3^{2}} + \frac{1}{4^{2}}} + . . . . n terms

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . . u p t o \infty = \frac{π^{2}}{6}

1 - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . . u p t o \infty

1

n

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + . . . . . . . .

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