Question

If n is a positive integer, show that $(p+q{)}^n-(n-1\left)pq\right(p+q{)}^{n-2}+\frac{(n-2)(n-3)}{\lfloor 2}{p}^2{q}^2(p-+q{)}^{n-4}-......$ is equal to $\frac{p^{n+1}-q^{n+1}}{p-q}$.

Answer 1

Let a series be;-

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

By the binomial theorem we see that

$1,n-1,\frac{(n-2)(n-3)}{2!},....$ are the co efficients of $x^n,x^{n-2},x^{n-4},x^{n-6}.....$ in the expansion of

${(1-x)}^{-1},{(1-x)}^{-2},{(1-x)}^{-3},{(1-x)}^{-4},.....$

Hence the sum required is equal to co efficient of $x^n$ in

$=\frac{1}{1-bx}-\frac{a{x}^2}{{(1-bx)}^2}+\frac{a^2{x}^4}{{(1-bx)}^3}-....$

$x^n$ in $S_{\infty }=\frac{1}{1-bx}÷(1+\frac{a{x}^2}{1-bx})$

$=\frac{1}{1-bx+a{x}^2}$

Hence for the series in question we put

$a=pq,b=p+q$

$\therefore S=\frac{1}{1-(p+q)x+pq{x}^2}$

$=\frac{1}{p-q}\{\frac{p}{1-px}-\frac{q}{1-qx}\}$

$\therefore$ sum is co efficient of $x^n$ in $S$

=coefficient of $x^n$ in $\frac{1}{p-q}\{p{(1-px)}^{-1}-q{(1-qx)}^{-1}\}$

$=\frac{p.p^n-q.q^n}{p-q}$

$=\frac{p^{n+1}-q^{n+1}}{p-q}$