Now Thats What we call a Summation!

Algebra Level 4

∑ r = 1 ∞ r 3 + ( r 2 + 1 ) 2 ( r 4 + r 2 + 1 ) ( r 2 + r ) = a b \large \sum^\infty_{r=1} \frac{r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)} = \frac ab

The equation above holds true for coprime positive integers a a and b b . Find a + b a+b .


The answer is 5.

Md Zuhair by Md Zuhair , 20, India

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2 solutions

Chew-Seong Cheong
Mar 10, 2018

S = ∑ r = 1 ∞ r 3 + ( r 2 + 1 ) 2 ( r 4 + r 2 + 1 ) ( r 2 + r ) = ∑ r = 1 ∞ r 4 + r 3 + 2 r 2 + 1 ( r 4 + r 2 + 1 ) ( r 2 + r ) = ∑ r = 1 ∞ ( 1 r ( r + 1 ) + r r 4 + r 2 + 1 ) = ∑ r = 1 ∞ ( 1 r − 1 r + 1 + r ( r 2 − r + 1 ) ( r 2 + r + 1 ) ) = 1 1 + ∑ r = 1 ∞ 1 2 ( 1 r 2 − r + 1 − 1 r 2 + r + 1 ) Let s = r + 1 = 1 + 1 2 ( ∑ r = 1 ∞ 1 r 2 − r + 1 − ∑ s = 2 ∞ 1 s 2 − s + 1 ) ⟹ r 2 + r + 1 = s 2 − s + 1 = 1 + 1 2 ( 1 ) = 3 2 \begin{aligned} S & = \sum_{r=1}^\infty \frac {r^3+(r^2+1)^2}{(r^4+r^2+1)(r^2+r)} \\ & = \sum_{r=1}^\infty \frac {r^4+r^3+2r^2+1}{(r^4+r^2+1)(r^2+r)} \\ & = \sum_{r=1}^\infty \left(\frac 1{r(r+1)} + \frac r{r^4+r^2+1}\right) \\ & = \sum_{r=1}^\infty \left(\frac 1r - \frac 1{r+1} + \frac r{(r^2-r+1)(r^2+r+1)}\right) \\ & = \frac 11 + \sum_{r=1}^\infty \frac 12 \left(\frac 1{r^2-r+1} -{\color{#3D99F6}\frac 1{r^2+r+1}} \right) & \small \color{#3D99F6} \text{Let }s = r+1 \\ & = 1 + \frac 12 \left(\sum_{r=1}^\infty \frac 1{r^2-r+1} -{\color{#3D99F6} \sum_{\color{#D61F06}s=2}^\infty \frac 1{s^2-s+1}} \right) & \small \color{#3D99F6} \implies r^2+r+1 = s^2 - s + 1 \\ & = 1 + \frac 12 (1) = \frac 32 \end{aligned}

Therefore, a + b = 3 + 2 = 5 a+b = 3+2 = \boxed{5} .

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Same here.

Ashutosh Sharma - 3 years, 3 months ago
Rohit Joshi
Apr 20, 2018

Llet r=1 Then equation be 9/6 Simplify then we get 3/2 which is a= 3 and b =2 So a+b=5

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