Exponential And Polynomials In One Sum?
The above equation is true for some coprime positive integers and . Find ?
The answer is 21.
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1 solution
Relevant wiki: Telescoping Series - Sum
\begin{aligned} S & = \sum_{n=2}^\infty \frac {n^4+3n^2+10n+10}{2^n(n^4+4)} \\ & = \sum_{n=2}^\infty \left(\frac 1{2^n} + \frac {3n^2+10n+6}{2^n\color{#3D99F6}{(n^4+4)}} \right) \quad \quad \small \color{#3D99F6}{\text{Note that }n^4+4 = (n^2+2n+2)(n^2-2n+2)} \\ & = \sum_{n=2}^\infty \left(\frac 1{2^n} + \frac {4(n^2+2n+2)-(n^2-2n+2)}{2^n\color{#3D99F6}{(n^2+2n+2)(n^2-2n+2)}} \right) \\ & = \sum_{n=2}^\infty \frac 1{2^n} + 4 \sum_{n=2}^\infty \frac 1{2^n(n^2-2n+2)} - \sum_{n=\color{#D61F06}{2}}^\infty \frac 1{2^\color{#D61F06}{n}(\color{#D61F06}{n}^2+2\color{#D61F06}{n}+2)} \\ & = \sum_{n=2}^\infty \frac 1{2^n} + 4 \sum_{n=2}^\infty \frac 1{2^n(n^2-2n+2)} - \sum_{n=\color{#D61F06}{4}}^\infty \frac 1{2^\color{#D61F06}{n-2}(\color{#D61F06}{(n-2)}^2+2\color{#D61F06}{(n-2)}+2)} \\ & = \sum_{n=2}^\infty \frac 1{2^n} + 4 \sum_{n=2}^\infty \frac 1{2^n(n^2-2n+2)} - \color{#D61F06}{4} \sum_ {n= \color{#D61F06}{4}}^\infty \frac 1{2^\color{#D61F06}{n}\color{#D61F06}{(n^2-2n+2)}} \\ & = \sum_{n=2}^\infty \frac 1{2^n} + \color{#D61F06}{4}\sum_{n= \color{#D61F06}{2}}^\color{#D61F06}{3} \frac 1{2^\color{#D61F06}{n}\color{#D61F06}{(n^2-2n+2)}} \\ & = \frac 1{2^2} \left(\frac 1{1-\frac 12} \right) + 4 \left(\frac 1{4(4-4+2)} + \frac 1{8(9-6+2)} \right) \\ & = \frac 12 + \frac 12 + \frac 1{10} = \frac {11}{10} \end{aligned}
⟹ a + b = 1 1 + 1 0 = 2 1