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

Answer: 7

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