Question

If $b=a+1$, and n is a positive integer, find the value of $b^n-(n-1)a{b}^{n-2}+\frac{(n-2)(n-3)}{\lfloor 2}{a}^2{b}^{n-4}-\frac{(n-3)(n-4)(n-5)}{\lfloor 3}{a}^3{b}^{n-6}+....$

Answer 1

By the Binomial Theorem, we see that
$1,n-1,\frac{(n-3)(n-2)}{\lfloor 2},\frac{(n-5)(n-4)(n-3)}{\lfloor 3},...$
are the coefficients of $x^n,x^{n-2},x^{n-4},x^{x-6},....$ in the expansions of $(1-x{)}^{-1},(1-x{)}^{-2},(1-x{)}^{-3},(1-x{)}^{-4},....$ respectively. Hence the sum required is equal to the coefficient of $x^n$ in the expansion of the series
$\frac{1}{1-bx}-\frac{a{x}^2}{(1-bx{)}^2}+\frac{a^2{x}^4}{(1-bx{)}^3}-\frac{a^3{x}^6}{(1-bx{)}^4}+...$,
and although the given expression consists only of a finite number of terms, this series may be considered to extend to infinity.
But the sum of the series $=\frac{1}{1-bx}÷(1+\frac{a{x}^2}{1-bx})=\frac{1}{1-bx+a{x}^2}$
$=\frac{1}{1-(a+1)x+a{x}^2},$ since $b=a+1$.
Hence the given series $=$coefficient of $x^n$ in $\frac{1}{(1-x)(1-ax)}$
$=$coefficient of $x^n$ in $\frac{1}{a-1}\left(\frac{a}{1-ax}-\frac{1}{1-x}\right)$
$=\frac{a^{n+1}-1}{a-1}$.