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Question 24
Find the sum $$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}} +....+ \sqrt{1+\frac{1}{2007^2}+\frac{1}{2008^2}}$$
Solution
Consider the first term:
$$\sqrt{1+1/1^2+1/2^2}$$ = $$\sqrt{9/4}$$ = 3/2
Second term: $$\sqrt{1+1/2^2+1/3^2}$$ = $$\sqrt{49/36}$$ = 7/6
First term + Second term = 3/2 + 7/6 = (9+7)/6 = 16/6 = 8/3 = 3 - 1/3
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Required sum = 2008 - 1/2008
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