WMC 2018 Senior Problem 31

Level 2

Evaluate 1 + 1 1 2 + 1 2 2 + 1 + 1 2 2 + 1 3 2 + 1 + 1 3 2 + 1 4 2 + ⋯ + 1 + 1 2017 2 + 1 2018 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}}+\cdots+\sqrt{1+\frac{1}{{2017}^2}+\frac{1}{{2018}^2}}

2017 − 1 2018 2017-\frac{1}{2018} 2017 − 1 2017 2017-\frac{1}{2017} 2018 − 1 2018 2018-\frac{1}{2018} 2018 − 1 2017 2018-\frac{1}{2017} 2018 − 1 2019 2018-\frac{1}{2019}

Tiger Ang by Tiger Ang , 20, Malaysia

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1 solution

X X
Oct 20, 2018

1 + 1 n 2 + 1 ( n + 1 ) 2 \space\space\space\space\sqrt{1+\dfrac1{n^2}+\dfrac1{(n+1)^2}}

= n 2 ( n + 1 ) 2 + n 2 + ( n + 1 ) 2 n 2 ( n + 1 ) 2 =\sqrt{\dfrac{n^2(n+1)^2+n^2+(n+1)^2}{n^2(n+1)^2}}

= n 4 + 2 n 3 + 3 n 2 + 2 n 2 + 1 n 2 ( n + 1 ) 2 =\sqrt{\dfrac{n^4+2n^3+3n^2+2n^2+1}{n^2(n+1)^2}}

= ( n 2 + n + 1 ) 2 n 2 ( n + 1 ) 2 =\sqrt{\dfrac{(n^2+n+1)^2}{n^2(n+1)^2}}

= n 2 + n + 1 n ( n + 1 ) =\dfrac{n^2+n+1}{n(n+1)}

= 1 + 1 n ( n + 1 ) =1+\dfrac1{n(n+1)}

= 1 + 1 n − 1 n + 1 =1+\dfrac1n-\dfrac1{n+1}

So, the expression becomes

( 1 + 1 1 − 1 2 ) + ( 1 + 1 2 − 1 3 ) + ( 1 + 1 3 − 1 4 ) + . . . ( 1 + 1 2017 − 1 2018 ) = 2018 − 1 2018 (1+\frac11-\frac12)+(1+\frac12-\frac13)+(1+\frac13-\frac14)+...(1+\frac1{2017}-\frac1{2018})=2018-\frac1{2018}

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