Partial fraction challenge

Algebra Level 5

S = 1 4 ( 1 ) 4 + 1 + 2 4 ( 2 ) 4 + 1 + 3 4 ( 3 ) 4 + 1 + ⋯ + 2016 4 ( 2016 ) 4 + 1 S=\frac{1}{4(1)^{4}+1}+\frac{2}{4(2)^{4}+1}+\frac{3}{4(3)^{4}+1}+\dots+\frac{2016}{4(2016)^{4}+1}

If the sum above S = 2017 p 2017 q + 1 S = \dfrac{2017p}{2017q+1} , where p p and q q are positive integers , find p + q p+q .


The answer is 5040.

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Tommy Li by Tommy Li , 22, Hong Kong

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2 solutions

Chew-Seong Cheong
Jun 22, 2016

S = ∑ n = 1 2016 n 4 n 4 + 1 = ∑ n = 1 2016 n ( 2 n 2 − 2 n + 1 ) ( 2 n 2 + 2 n + 1 ) = 1 4 ∑ n = 1 2016 ( 1 2 n 2 − 2 n + 1 − 1 2 n 2 + 2 n + 1 ) = 1 4 ( ∑ n = 1 2016 1 2 n 2 − 2 n + 1 − ∑ n = 1 2016 1 2 n 2 + 2 n + 1 ) Let n = m − 1 = 1 4 ( ∑ n = 1 2016 1 2 n 2 − 2 n + 1 − ∑ m = 2 2017 1 2 ( m − 1 ) 2 + 2 ( m − 1 ) + 1 ) = 1 4 ( ∑ n = 1 2016 1 2 n 2 − 2 n + 1 − ∑ m = 2 2017 1 2 m 2 − 2 m + 1 ) Putting m = n = 1 4 ( ∑ n = 1 2016 1 2 n 2 − 2 n + 1 − ∑ n = 2 2017 1 2 n 2 − 2 n + 1 ) = 1 4 ( 1 − 1 2 ( 2017 ) 2 − 2 ( 2017 ) + 1 ) = 1 4 ( 1 − 1 2017 ⋅ 4032 + 1 ) = 1 4 ( 2017 ⋅ 4032 2017 ⋅ 4032 + 1 ) = 2017 ⋅ 1008 2017 ⋅ 4032 + 1 \begin{aligned} S & = \sum_{n=1}^{2016} \frac n{4n^4+1} \\ & = \sum_{n=1}^{2016} \frac n{(2n^2-2n+1)(2n^2+2n+1)} \\ & = \frac 14 \sum_{n=1} ^{2016} \left( \frac 1{2n^2-2n+1} - \frac 1{2n^2+2n+1} \right) \\ & = \frac 14 \left( \sum_{n=1} ^{2016} \frac 1{2n^2-2n+1} - \sum_{\color{#3D99F6}{n=1}} ^{\color{#3D99F6}{2016}} \frac 1{2\color{#3D99F6}{n}^2+2\color{#3D99F6}{n}+1} \right) \quad \quad \small \color{#3D99F6}{\text{Let }n=m-1} \\ & = \frac 14 \left( \sum_{n=1} ^{2016} \frac 1{2n^2-2n+1} - \sum_{\color{#3D99F6}{m=2}} ^{\color{#3D99F6}{2017}} \frac 1{2\color{#3D99F6}{(m-1)}^2+2\color{#3D99F6}{(m-1)}+1} \right) \\ & = \frac 14 \left( \sum_{n=1} ^{2016} \frac 1{2n^2-2n+1} - \sum_{\color{#3D99F6}{m=2}} ^{\color{#3D99F6}{2017}} \frac 1{2\color{#3D99F6}{m}^2-2\color{#3D99F6}{m}+1} \right) \quad \quad \small \color{#3D99F6}{\text{Putting }m=n} \\ & = \frac 14 \left( \sum_{n=1} ^{2016} \frac 1{2n^2-2n+1} - \sum_{n=2} ^{2017} \frac 1{2n^2-2n+1} \right) \\ & = \frac 14 \left( 1 - \frac 1{2(2017)^2-2(2017)+1} \right) \\ & = \frac 14 \left( 1 - \frac 1{2017 \cdot 4032 +1} \right) \\ & = \frac 14 \left(\frac {2017 \cdot 4032}{2017 \cdot 4032 +1} \right) \\ & = \frac {2017 \cdot 1008}{2017 \cdot 4032 +1} \end{aligned}

⟹ p + q = 1008 + 4032 = 5040 \implies p+q = 1008 + 4032 = \boxed{5040}

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Thanks for your solution!

Tommy Li - 4 years, 11 months ago

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Your hint completely gave it away...

A Former Brilliant Member - 4 years, 11 months ago

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Oh, I should remove the hint to higher the difficulty of this problem.

Tommy Li - 4 years, 11 months ago

even i used the sophie germain identity

abhishek alva - 4 years, 3 months ago
Tommy Li
Jun 22, 2016

x 4 x 4 + 1 ≡ A 2 x 2 − 2 x + 1 + B 2 x 2 + 2 x + 1 ≡ A 2 x ( x − 1 ) + 1 + B 2 x ( x + 1 ) + 1 \large \frac{x}{4x^{4}+1}\equiv\frac{A}{2x^{2}-2x+1}+\frac{B}{2x^{2}+2x+1}\equiv\frac{A}{2x(x-1)+1}+\frac{B}{2x(x+1)+1}

x ≡ A ( 2 x 2 + 2 x + 1 ) + B ( 2 x 2 − 2 x + 1 ) x\equiv A(2x^{2}+2x+1)+B(2x^{2}-2x+1)

⇒ A = 1 4 , B = − 1 4 \Rightarrow A=\frac{1}{4} , B=-\frac{1}{4}

1 4 ( 1 ) 4 + 1 + 2 4 ( 2 ) 4 + 1 + 3 4 ( 3 ) 4 + 1 + ⋯ + 2016 4 ( 2016 ) 4 + 1 \frac{1}{4(1)^{4}+1}+\frac{2}{4(2)^{4}+1}+\frac{3}{4(3)^{4}+1}+\dots+\frac{2016}{4(2016)^{4}+1}

= 1 4 ( ( 1 2 ( 1 ) ( 1 − 1 ) + 1 − 1 2 ( 1 ) ( 1 + 1 ) + 1 ) + ( 1 2 ( 2 ) ( 2 − 1 ) + 1 − 1 2 ( 2 ) ( 2 + 1 ) + 1 ) + ( 1 2 ( 3 ) ( 3 − 1 ) + 1 − 1 2 ( 3 ) ( 3 + 1 ) + 1 ) + ⋯ + ( 1 2 ( 2016 ) ( 2016 − 1 ) + 1 − 1 2 ( 2016 ) ( 2016 + 1 ) + 1 ) ) =\frac{1}{4}((\frac{1}{2(1)(1-1)+1}-\frac{1}{2(1)(1+1)+1})+(\frac{1}{2(2)(2-1)+1}-\frac{1}{2(2)(2+1)+1})+(\frac{1}{2(3)(3-1)+1}-\frac{1}{2(3)(3+1)+1})+\dots+(\frac{1}{2(2016)(2016-1)+1}-\frac{1}{2(2016)(2016+1)+1}))

= 1 4 ( ( 1 1 − 1 5 ) + ( 1 5 − 1 13 ) + ( 1 13 − 1 25 ) + ⋯ + 1 2 ( 2016 ) ( 2016 − 1 ) + 1 − 1 2 ( 2016 ) ( 2016 + 1 ) + 1 ) ) =\frac{1}{4}((\frac{1}{1}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{13})+(\frac{1}{13}-\frac{1}{25})+\dots+\frac{1}{2(2016)(2016-1)+1}-\frac{1}{2(2016)(2016+1)+1}))

= 1 4 ( 1 − 1 4032 ( 2017 ) + 1 ) =\frac{1}{4}(1-\frac{1}{4032(2017)+1})

= 1 4 ( 4032 ( 2017 ) + 1 − 1 4032 ( 2017 ) + 1 ) =\frac{1}{4}(\frac{4032(2017)+1-1}{4032(2017)+1})

= 1008 ( 2017 ) 4032 ( 2017 ) + 1 =\frac{1008(2017)}{4032(2017)+1}

⇒ p + q = 1008 + 4032 = 5040 \Rightarrow p+q=1008+4032=5040

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