Question

If $n$ is a positive integer, show that
(1) $n^{n+1}-n{(n-1)}^{n+1}+\frac{n(n-1)}{2!}{(n-2)}^{n+1}-\cdots =\frac{1}{2}n(n+1)!$;
(2) $n^n-(n+1){(n-1)}^n+\frac{(n+1)n}{2!}{(n-2)}^n-\cdots =1$;
the series in each case being extended to $n$ terms; and
(3) $1^n-n{2}^n+\frac{n(n-1)}{1\cdot 2}{3}^n-\cdots ={(-1)}^{n}n!$;
(4) ${(n+p)}^n-n{(n+p-1)}^n+\frac{n(n-1)}{2!}{(n+p-2)}^n-\cdots =n!$;
the series in the last two cases being extended to $n+1$ terms.

Answer 1

If n is a positive , Show that

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

Since , We have

$\frac{n^{n+1}}{(n+1)!}-\frac{n(n-1{)}^{n+1}}{(n+1)!}+\frac{n(n-1)}{2!}.\frac{(n-2{)}^{n+1}}{(n+1)!}-......tonterms$ = The Coefficient of $x^{n+1}$'

in ${(x+\frac{x^2}{2!}+\frac{x^3}{3!}+........)}^n=\frac{1}{2}n$

Therefore , On Solving this we get

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

2) To Show that

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

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

$=x^{n+1}$ + Terms Containing Higher Power of x

On Expanding this ,

$e^{nx}-(n+1)e^{(n-1)x}+\frac{n(n+1)}{1.2}{e}^{(n-2)x}-.....to(n+2)terms=e^{-x}(x^{n+1}+.......)$

Therefore , Coefficient of $x^n$ in the series is

$\frac{n^2}{n!}-(n+1)\frac{(n-1{)}^n}{n!}+\frac{n.(n+1)}{1.2}.\frac{(n-2{)}^n}{n!}-....$ to n terms

$+(-1{)}^{n+1}\frac{(-1{)}^n}{n!}=0$

On Multiplying by $n!$ and Simplifying we get

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

3 ) To Show that

$1^n-n{2}^n+\frac{n(n-1)}{1.2}{3}^n-....=(-1{)}^{n}n!$

We have ,

$e^x(1-e^x{)}^n=e^x-n{e}^{2x}+\frac{n(n-1)}{1.2}{e}^{3x}......to(n+1)terms$

$\therefore$

The Coefficient of $x^n$ in the Expression is

$\frac{1}{n!}-n\frac{2^n}{n!}+\frac{n(n-1)}{1.2}.\frac{3^n}{n!}-......nterms$

Thus , the Coefficient of $x^n$ is $(-1{)}^n$

Equate the two Coefficients , Multiply up by $n!$

Finally we have

$1^n-n{2}^n+\frac{n(n-1)}{1.2}{3}^n-....=(-1{)}^{n}n!$

4 ) To Show that ,

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

We have ,

$e^{px}{(e^x-1)}^n=e^{px}(e^{nx}-n{e}^{(n-1)x}+\frac{n(n-1)}{1.2}{e}^{(n-2)x}-......)$

$=(e^{px}.e^{nx}-n{e}^{px}.e^{(n-1)x}+\frac{n(n-1)}{1.2}{e}^{(n-2)x}.e^{px}-......)$

$=(e^{(p+n)}-n{e}^{(p+n-1)x}+\frac{n(n-1)}{1.2}{e}^{(p+n-2)x}-......)$

To Equate the Coefficient of $x^n$ , we have

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