Question

$n^n-n(n-1{)}^n+\frac{n(n-1)}{2!}(n-2{)}^n-...\infty =\frac{3}{a}\times n!$.

Find $a$

Answer 1

We have

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

$\Rightarrow$
$e^x-1=x(\frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+...)$

$\therefore$ $(e^x-1{)}^n=x^n{(\frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+...)}^n$

$=x^n+$ terms containing higher power of x ...(1)

Using binomial theorem, we get

$(e^x-1{)}^n=e^{nx}{-}^n{C}_1{e}^{(n-1)x}{+}^n{C}_2{e}^{(n-2)x}-...\infty$

$=[1+(nx)+\frac{(nx{)}^2}{2!}+...+\frac{(nx{)}^n}{n!}+...]-n[1+(n-1)x+\frac{(n-1{)}^2{x}^2}{2!}+...+\frac{(n-1{)}^n{x}^n}{n!}+...]+\frac{n(n-1)}{2!}[1+(n-2)x+\frac{(n-2{)}^2{x}^2}{2!}+...+\frac{(n-2{)}^n{x}^n}{n!}+...]+.....$

Above is eq $\left(2\right)$.

Equating the coefficient of $x^n$ in (1) & (2), we get

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

or $n^n-n(n-1{)}^n+\frac{n(n-1)}{2!}(n-2{)}^n-...=n!$

$\therefore a=3$