Let $$\sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$ and $e=\sum_\limits{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}$ Then $\mathrm{a}^{2}-\mathrm{b}+\mathrm{c}$ is equal to ____________.
Answer (integer)
26
Solution
<p>$\sum\limits_{n = 0}^\infty {{{{n^3}(2n!) + (2n - 1)(n!)} \over {n!\,.\,(2n)!}}}$</p>
<p>$= \sum\limits_{n = 0}^\infty {{{{n^3}} \over {n!}} + {{2n - 1} \over {2n!}}}$</p>
<p>$$ = \sum\limits_{n = 0}^\infty {{3 \over {(n - 2)!}} + {1 \over {(n - 3)!}} + {1 \over {(n - 1)!}} + {1 \over {(2n - 1)!}} - {1 \over {(2n)!}}} $$</p>
<p>$= 3e + e + e - {1 \over e}$</p>
<p>$= 5e - {1 \over e}$</p>
<p>$\therefore$ $a = 5,b = - 1,c = 0$</p>
<p>$\therefore$ ${a^2} - b + c = 26$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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