Multiple Partial Fractions

Algebra Level 4

x 3 + x 2 + x + k ( x + k ) 2 ( x 2 + k 2 ) \large \frac{x^3+x^2+x+k}{(x+k)^2 (x^2+k^2)}

Given that k ∈ Z + k\in\mathbb{Z}^+ is a constant and the expression above decomposes into partial fractions of

C k , 1 x + k + C k , 2 ( x + k ) 2 + C k , 3 x + C k , 4 x 2 + k 2 \frac{C_{k,1}}{x+k}+\frac{C_{k,2}}{(x+k)^2}+\frac{C_{k,3}x+C_{k,4}}{x^2+k^2}

where C k , 1 C_{k,1} , C k , 2 C_{k,2} , C k , 3 C_{k,3} , and C k , 4 C_{k,4} are constants. Determine the value of ∑ k = 2 50 C k , 2 \displaystyle\sum_{k=2}^{50} {C_{k,2}} .

− 1225 4 -\frac{1225}{4} 1225 1225 − 1225 2 -\frac{1225}{2} 1225 4 \frac{1225}{4} − 1225 -1225 1225 2 \frac{1225}{2}

Wee Xian Bin by Wee Xian Bin , 25, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Sep 26, 2017

x 3 + x 2 + x + k ( x + k ) 2 ( x 2 + k 2 ) = C k , 1 x + k + C k , 2 ( x + k ) 2 + C k , 3 x + C k , 4 x 2 + k 2 Multiply both sides by ( x + k ) 2 ( x 2 + k 2 ) x 3 + x 2 + x + k = C k , 1 ( x + k ) ( x 2 + k 2 ) + C k , 2 ( x 2 + k 2 ) + ( C k , 3 x + C k , 4 ) ( x + k ) 2 Puting x = − k − k 3 + k 2 − k + k = C k , 2 ( k 2 + k 2 ) ⟹ C k , 2 = − k − 1 2 \begin{aligned} \frac{x^3+x^2+x+k}{(x+k)^2 (x^2+k^2)} & = \frac{C_{k,1}}{x+k}+\frac{C_{k,2}}{(x+k)^2}+\frac{C_{k,3}x+C_{k,4}}{x^2+k^2} & \small \color{#3D99F6} \text{Multiply both sides by }(x+k)^2 (x^2+k^2) \\ x^3+x^2+x+k & = C_{k,1}(x+k)(x^2+k^2)+C_{k,2}(x^2+k^2)+ \left(C_{k,3}x+C_{k,4}\right)(x+k)^2 & \small \color{#3D99F6} \text{Puting }x = - k \\ -k^3+k^2-k+k & = C_{k,2}(k^2+k^2) \\ \implies C_{k,2} & = - \frac {k-1}2 \end{aligned}

Therefore,

S = ∑ k = 2 50 C k , 2 = − 1 2 ∑ k = 2 50 ( k − 1 ) = − 1 2 ∑ k = 1 49 k − 1 2 × 49 × 50 2 = − 1225 2 \begin{aligned} S & = \sum_{k=2}^{50} C_{k,2} \\ & = - \frac 12 \sum_{\color{#3D99F6}k=2}^{\color{#3D99F6}50} {\color{#3D99F6}(k-1)} \\ & = - \frac 12 \sum_{\color{#D61F06} k=1}^{\color{#D61F06}49} {\color{#D61F06}k} \\ & - \frac 12 \times \frac {49\times 50}2 \\ & = \boxed{-\dfrac {1225}2} \end{aligned}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...