Trigo Roller Coasters

Geometry Level 3

Given that S n = ∑ k = 1 n tan ⁡ − 1 1 k ( k + 1 ) + 1 \displaystyle S_{n}=\sum_{k=1}^{n} \tan^{-1} \cfrac{1}{k(k+1)+1} for positive integers n ∈ N n\in N , evaluate 1 + ∑ n = 1 62 1 + tan ⁡ S n 1 − tan ⁡ S n \displaystyle 1+\sum_{n=1}^{62}\cfrac{1+ \tan S_{n}}{1- \tan S_{n}} .

This is part of the set Fun With Problem-Solving .


The answer is 2016.

Donglin Loo by donglin loo , 22, Singapore

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1 solution

Donglin Loo
May 25, 2018

S n = ∑ k = 1 n tan ⁡ − 1 1 k ( k + 1 ) + 1 S_{n}=\sum_{k=1}^{n} \tan^{-1} \cfrac{1}{k(k+1)+1} .

S n = ∑ k = 1 n tan ⁡ − 1 1 k ( k + 1 ) 1 + 1 k ( k + 1 ) S_{n}=\sum_{k=1}^{n} \tan^{-1} \cfrac{\cfrac{1}{k(k+1)}}{1+\cfrac{1}{k(k+1)}} .

S n = ∑ k = 1 n tan ⁡ − 1 1 k − 1 k + 1 1 + 1 k ( k + 1 ) S_{n}=\sum_{k=1}^{n} \tan^{-1} \cfrac{\cfrac{1}{k}-\cfrac{1}{k+1}}{1+\cfrac{1}{k(k+1)}} .

S n = ∑ k = 1 n ( tan ⁡ − 1 1 k − tan ⁡ − 1 1 k + 1 ) S_{n}=\sum_{k=1}^{n} (\tan^{-1}\cfrac{1}{k}-\tan^{-1}\cfrac{1}{k+1}) .

S n = tan ⁡ − 1 1 1 − tan ⁡ − 1 1 2 + tan ⁡ − 1 1 2 − tan ⁡ − 1 1 3 + . . . + tan ⁡ − 1 1 n − 1 − tan ⁡ − 1 1 n + tan ⁡ − 1 1 n − tan ⁡ − 1 1 n + 1 S_{n}=\tan^{-1}\cfrac{1}{1}-\tan^{-1}\cfrac{1}{2}+\tan^{-1}\cfrac{1}{2}-\tan^{-1}\cfrac{1}{3}+...+\tan^{-1}\cfrac{1}{n-1}-\tan^{-1}\cfrac{1}{n}+\tan^{-1}\cfrac{1}{n}-\tan^{-1}\cfrac{1}{n+1}

S n = tan ⁡ − 1 1 − tan ⁡ − 1 1 n + 1 = π 4 − t a n − 1 1 n + 1 S_{n}=\tan^{-1}{1}-\tan^{-1}\cfrac{1}{n+1}=\cfrac{\pi}{4}-tan^{-1}\cfrac{1}{n+1}

∴ 1 + ∑ n = 1 62 1 + tan ⁡ S n 1 − tan ⁡ S n \therefore 1+\sum_{n=1}^{62}\cfrac{1+\tan S_{n}}{1-\tan S_{n}}

= 1 + ∑ n = 1 62 tan ⁡ ( π 4 + S n =1+\sum_{n=1}^{62}\tan (\cfrac{\pi}{4}+S_{n} )

= 1 + ∑ n = 1 62 tan ⁡ ( π 4 + π 4 − t a n − 1 1 n + 1 ) =1+\sum_{n=1}^{62}\tan (\cfrac{\pi}{4}+\cfrac{\pi}{4}-tan^{-1}\cfrac{1}{n+1})

= 1 + ∑ n = 1 62 tan ⁡ ( π 2 − t a n − 1 1 n + 1 ) =1+\sum_{n=1}^{62}\tan (\cfrac{\pi}{2}-tan^{-1}\cfrac{1}{n+1})

= 1 + ∑ n = 1 62 cot ⁡ ( t a n − 1 1 n + 1 ) =1+\sum_{n=1}^{62}\cot (tan^{-1}\cfrac{1}{n+1})

= 1 + ∑ n = 1 62 1 tan ⁡ ( t a n − 1 1 n + 1 ) =1+\sum_{n=1}^{62}\cfrac{1}{\tan (tan^{-1}\cfrac{1}{n+1})}

= 1 + ∑ n = 1 62 1 1 n + 1 =1+\sum_{n=1}^{62}\cfrac{1}{\cfrac{1}{n+1}}

= 1 + ∑ n = 1 62 ( n + 1 ) =1+\sum_{n=1}^{62}(n+1)

= 1 + ( 2 + 3 + . . . + 63 ) =1+(2+3+...+63)

= 63 ⋅ 64 2 =\cfrac{63\cdot64}{2}

= 2016 =2016

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