Sequences & Series
Infinite series involving factorials and exponential
nta_pyq_2023_jan
Grade 11
Question:
Let $\displaystyle\sum_{n=0}^{\infty} \dfrac{n^3((2n)!) + (2n-1)(n!)}{(n!)((2n)!)} = ae + \dfrac{b}{e} + c$, where $a, b, c \in \mathbb{Z}$ and $e = \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{n!}$. Then $a^2 - b + c$ is equal to ______.
Step-by-Step Solution
Key Concept: Split the sum into $\sum \frac{n^3}{n!} + \sum \frac{2n-1}{(2n)!}$. The first part involves $e$ via $\sum \frac{n^k}{n!}$. The second splits into $\frac{1}{2}(e - 1/e) - \frac{1}{2}(e + 1/e)$.
The series evaluates to $5e - \frac{1}{e}$, so $a=5$, $b=-1$, $c=0$. Therefore $a^2 - b + c = 25 + 1 + 0 = 26$.
Correct Answer: 26