Series

Algebra Level 3

1 + 1 1 2 + 1 2 2 + 1 + 1 2 2 + 1 3 2 + ⋯ + 1 + 1 9 9 2 + 1 10 0 2 = ? \sqrt{ 1+ \dfrac1{1^2} +\dfrac1{2^2}} + \sqrt{ 1+ \dfrac1{2^2} +\dfrac1{3^2}} + \cdots + \sqrt{ 1+ \dfrac1{99^2} +\dfrac1{100^2}} = \, ?

100 100 100 − 1 100 100-\frac1{100} 99 99 99 − 1 99 99-\frac1{99}

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Sridhar Sri by Sridhar Sri , 25, India

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1 solution

Kay Xspre
Jan 6, 2016

This question asked us to find the value of ∑ n = 1 99 1 + 1 n 2 + 1 ( n + 1 ) 2 \sum_{n=1}^{99} \sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}} We start by simplifying the roots, which gives 1 + 2 n ( n + 1 ) + 1 ( n ( n + 1 ) ) 2 = ( n ( n + 1 ) ) + 1 ) 2 ( n ( n + 1 ) ) 2 = 1 + 1 n ( n + 1 ) \sqrt{1+\frac{2n(n+1)+1}{(n(n+1))^2}} = \sqrt{\frac{(n(n+1))+1)^2}{(n(n+1))^2}} = 1+\frac{1}{n(n+1)} Then we return to sigma, which gives ∑ n = 1 99 ( 1 + 1 n ( n + 1 ) ) = 99 + ( 1 − 1 100 ) = 100 − 1 100 = 99.99 \sum_{n=1}^{99} (1+\frac{1}{n(n+1)}) = 99+(1-\frac{1}{100}) = 100-\frac{1}{100} = 99.99

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