Question

Let $a_n=\underset{-1}{\overset{n}{\int }}\left(1+\frac{x}{2}+\frac{x^2}{2}+\frac{x^3}{3}+\dots \dots ..+\frac{x^{n-1}}{n}\right)dx$ for $n\in N$. Then the sum of all the elements of the set $\left\{n\in N:a_n\in (2,30)\right\}$ is ______

Answer 1

Answer: 5

$\underset{-1}{\overset{n}{\int }}\left(1+\frac{x}{2}+\frac{x^2}{3}+\dots +\frac{x^{n-1}}{n}\right)dx$
${\left[x+\frac{x^2}{2}+\frac{x^3}{3^2}+\dots +\frac{x^n}{n^2}\right]}_{-1}^n$
$\left(n+\frac{n^2}{2^2}+\frac{n^3}{3^2}+\dots +-\frac{n^n}{n^2}\right)$
$-\left(-1+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}+\dots +\frac{(-1{)}^n}{n^2}\right)$
$a_n=(n+1)+\frac{1}{2^2}\left(n^2-1\right)+\frac{1}{3^2}\left(n^3+1\right)$
$+\dots +\frac{1}{n^2}\left(n^n-(-1{)}^n\right)$
if $n=1\Rightarrow a_n=2\notin (2,30)$
if $n=2$
$\Rightarrow a_n=(2+1)+\frac{1}{2^2}\left(2^2-1\right)=3+\frac{3}{4}<30$
if $n=3$
$\Rightarrow a_n=(3+1)+\frac{1}{4}(8)+\frac{1}{9}(28)=11+\frac{28}{9}<30$
If $n=4$
$\Rightarrow a_n=(4+1)+\frac{1}{4}(16-1)+\frac{1}{9}(64+1)+\frac{1}{16}$
$=5+\frac{15}{4}+\frac{65}{9}+\frac{255}{16}>30$
Test $\{2,3\}$ sum of elements 5