Question

Let $a _{ n }=\int\limits_{-1}^{ n }\left(1+\frac{ x }{2}+\frac{ x ^2}{2}+\frac{ x ^3}{3}+\ldots \ldots +\frac{ x ^{ n -1}}{ n }\right) dx$ 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

The correct answer is 5.

$\int_{-1}^{n}(1+\frac{x}{2}+\frac{x^{2}}{3}+....+\frac{x^{n-1}}{n})dx$

$[x+\frac{x^{2}}{2}+\frac{x^{3}}{3^{2}}+...\frac{x^{n}}{n^{2}}]^{n}$

$(n+\frac{n^{2}}{2^{2}}+\frac{n^{3}}{3^{2}}+...+-\frac{n^{n}}{n^{2}})$

$-(-1+\frac{1}{2^{2}}-\frac{1}{3^{2}}+\frac{1}{4^{2}}+...+\frac{(-1)^{n}}{n^{2}})$

$a_{n}=(n+1)+\frac{1}{2^{2}}(n^{2}-1)+\frac{1}{3^{2}}(n^{3}+1)+...+\frac{1}{n^{2}}(n^{n}-(-1)^{n})$

$If \; n=1\Rightarrow a_{n}=2\not{\epsilon }(2,30)$

If n = 2

$\Rightarrow a_{n}=(2+1)+\frac{1}{2^{2}}(2^{2}-1)=3+\frac{3}{4}< 30$

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