Question

If $\int e ^{ x }\left(\frac{1- x ^{ n }}{1- x }\right) dx = e ^{ x } \cdot P ( x )+ C$ where $n \in N$ and $C ^{\prime}$ is constant of integration and $P (0)$ $=620$ then find the value of $n$.

Answer 1

$ \int e ^{ x }\left(\frac{1- x ^{ n }}{1- x }\right) dx = e ^{ x } P ( x )+ C$
Let $P ( x )= a _0+ a _1 x + a _2 x ^2+\ldots \ldots .+ a _{ n -1} x ^{ n -1}$
$=\int e^x\left(1+x+x^2+\ldots \ldots+x^{n-1}\right) d x=e^x\left(a_0+a_1 x+a_2 x^2+\ldots \ldots+a_{n-1} x^{n-1}\right)+c $
$P(0)=a_0=620$
Differentiating both sides
$e ^{ x }\left(1+ x + x ^2+\ldots \ldots x ^{ n -1}\right)= e ^{ x }\left( a _1+2 a _2 x +3 a _3 x ^2\right. \left.+\ldots \ldots+( n -1)_{ n -1} x ^{ n -2}\right) $
$ +\left( a _0+ a _1 x + a _2 x ^2+\ldots \ldots+ a _{ n -1} x ^{ n -1}\right) e ^{ x }$
Comparing coefficient of same power of $x$
$a_{n-1}=1 $
$a_0=620$
$a_1+a_0=1 \Rightarrow a_1=-619 $
$a_1+2 a_2=1 \Rightarrow a_2=310$
$a_2+3 a_3=1 \Rightarrow a_3=-103 $
$a_3+4 a_4=1 \Rightarrow a_4=+26$
$a_4+5 a_5=1 \Rightarrow a_5=-5$
$a_5+6 a_6=1 \Rightarrow a_6=1 $
$\therefore n-1=6 \Rightarrow n=7$