Let a_n = _-1^n(1+x2+x^23++x^n-1n)dx for every nN.

Question

Let $a_n = \displaystyle\int_{-1}^{n}\left(1+\frac{x}2+\frac{x^2}3+\cdots+\frac{x^{n-1}}n\right)dx$ for every $n\in\mathbb{N}$. Then the sum of all the elements of the set $\{n\in\mathbb{N}: a_n\in(2,30)\}$ is\_\_\_

Answer: 5

Accepted Answer

$$a_n = \left[x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+\cdots+\frac{x^n}{n^2}\right]_{-1}^n = \sum_{k=1}^n\frac{n^k-(-1)^k}{k^2}$$

Computing successive values: $a_1=2$, $a_2=\dfrac92$, $a_3=\dfrac{100}9$, $a_4 > 31$.

Since $a_1=2\notin(2,30)$ (endpoint excluded) but $a_2,a_3\in(2,30)$ and $a_4\notin(2,30)$, the qualifying set is $\{2,3\}$, with sum $=5$.

Solutions · 1