Factoring a quartic

Algebra Level 4

If ∑ k = 1 20 k 2 − 1 2 k 4 + 1 4 = A B \sum\limits_{k=1}^{20}\frac{k^2-\frac12}{k^4+\frac14}=\frac{A}{B} , where A A and B B are relatively coprime positive integers., find A A


The answer is 800.

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Barack Clinton by Barack Clinton , 32, Switzerland

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1 solution

Barack Clinton
Feb 17, 2015

k 4 + 1 4 = ( k 2 − k + 1 2 ) ( k 2 + k + 1 2 ) k^4+\frac{1}{4}=\left(k^2-k+\frac12\right)\left(k^2+k+\frac12\right) Thus; ∑ k = 1 n k 2 − 1 2 k 4 + 1 4 = ∑ k = 1 n ( k − 1 2 k 2 − k + 1 2 − k + 1 2 k 2 + k + 1 2 ) = ∑ k = 1 n ( k − 1 2 k 2 − k + 1 2 − ( k + 1 ) − 1 2 ( k + 1 ) 2 − ( k + 1 ) + 1 2 ) \sum\limits_{k=1}^{n}\frac{k^2-\frac12}{k^4+\frac14}=\sum\limits_{k=1}^{n}\left(\frac{k-\frac12}{k^2-k+\frac12}-\frac{k+\frac12}{k^2+k+\frac12}\right)=\sum\limits_{k=1}^{n}\left(\frac{k-\frac12}{k^2-k+\frac12}-\frac{(k+1)-\frac12}{(k+1)^2-(k+1)+\frac12}\right)

Since the sum above is telescoping, we get the general formula 1 − 2 n + 1 2 n 2 + 2 n + 1 1-\frac{2n+1}{2n^2+2n+1}

For n = 20 n=20 it gives 800 841 ⇒ A = 800 \frac{800}{841}\Rightarrow A=800

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