Don't forget 2015

Algebra Level 5

k = 2 63 k 3 k k 6 1 \large \displaystyle \sum_{k=2}^{63} \dfrac{k^3-k}{k^6-1}\

If the value of the summation above is in the form p q \dfrac{p}{q} , where p p and q q are coprime positive integers, find p + q p+q .


The answer is 14114.

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2 solutions

Rishabh Jain
Jan 27, 2016

k = 2 63 k 3 k k 6 1 = k ( k 2 1 ) ( k 2 1 ) ( k 4 + k 2 + 1 ) \large \displaystyle \sum_{k=2}^{63} \dfrac{k^3-k}{k^6-1}=\dfrac{k(\color{#D61F06}{k^2-1})}{(\color{#D61F06}{k^2-1})(k^4+k^2+1)} = 1 2 k = 2 63 1 k 2 k + 1 1 k 2 + k + 1 ( T e l e s c o p i c S e r i e s ) =\dfrac{1}{2} \displaystyle \sum_{k=2}^{63} \dfrac{1}{k^2-k+1}- \dfrac{1}{k^2+k+1}~~\color{#20A900}{(Telescopic~Series)} = 1 2 ( 1 3 1 4033 ) =\dfrac{1}{2}(\dfrac{1}{3}-\dfrac{1}{4033}) = 2015 12099 =\dfrac{\color{goldenrod}{2015}}{12099} 2015 + 12099 = 14114 \Large 2015+12099=\boxed{\color{darkviolet}{\boxed{14114}}}

Rohit Udaiwal
Jan 27, 2016

k 3 k k 6 1 = k ( k 2 1 ) ( k 3 ) 2 1 = k ( k 1 ) ( k + 1 ) ( k 3 + 1 ) ( k 3 1 ) = k ( k 1 ) ( k + 1 ) ( k + 1 ) ( k 2 + 1 k ) ( k 1 ) ( k 2 + 1 + k ) = k ( k 2 + 1 k ) ( k 2 + 1 + k ) = 1 2 ( 1 k 2 + 1 k 1 k 2 + 1 + k ) \dfrac{k^3-k}{k^6-1}=\dfrac{k(k^2-1)}{(k^3)^2-1} =\dfrac{k(k-1)(k+1)}{(k^3+1)(k^3-1)} =\dfrac{k(k-1)(k+1)}{(k+1)(k^2+1-k)(k-1)(k^2+1+k)} \\ =\dfrac{k}{(k^2+1-k)(k^2+1+k)} \\ =\dfrac{1}{2}\left(\dfrac{1}{k^2+1-k}-\dfrac{1}{k^2+1+k}\right) k = 2 63 k 3 k k 6 1 = 1 2 n = 2 63 1 k 2 + 1 k 1 k 2 + 1 + k = 1 2 ( 1 3 1 7 + 1 7 1 13 + 1 13 1 6 3 2 + 1 + 63 ) = 1 2 ( 1 3 1 4033 ) = 1 2 4030 12099 = 2015 12099 \begin{aligned} \displaystyle \sum_{k=2}^{63} \dfrac{k^3-k}{k^6-1} & =\dfrac{1}{2}\displaystyle \sum_{n=2}^{63} \dfrac{1}{k^2+1-k}-\dfrac{1}{k^2+1+k} \\ & =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\ldots- \dfrac{1}{63^2+1+63}\right) \\ & =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{4033}\right) =\dfrac{1}{2}\cdot\dfrac{4030}{12099} \\ &=\boxed{\dfrac{2015}{12099}} \end{aligned}

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