Question

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

show that $1-(n-1)+\frac{(n-2)(n-2)}{\lfloor 2}-\frac{(n-3)(n-4)(n-5)}{\lfloor 3}+....=(-1{)}^n$.

Answer 1

By binomial theorem we see that,

$1,n-1,\frac{(n-3)(n-2)}{2!},\frac{(n-5)(n-4)(n-3)}{3!},...$ are the co efficient of $x^n,x^{n-2},x^{n-4},....$ in the expansion of ${(1-x)}^{-1},{(1-x)}^{-2},{(1-x)}^{-3},...$ respectively. Hence the sum required is equal to the co efficient of $x^n$ in the expansion of the series;-

$\frac{1}{1-bx}-\frac{a{x}^2}{{(1-bx)}^2}+\frac{a^2{x}^4}{{(1-bx)}^3}-\frac{a^3{x}^6}{{(1-bx)}^4}+....$

Where $a=1,b=\mathrm{2...}$

$\therefore$ The sum of the series;-

$S=\left(\frac{1}{1-bx}\right)÷(1+\frac{a{x}^2}{1-bx})$

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

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

$b=2,a=1$

$S=\frac{1}{1-2x+x^2}={(1-x)}^{-2}$

$S=$ co efficient of $x^n$ in ${(1-x)}^{-2}$

$=n+1$

By putting $a=1,b=1$ we get sum of second series;-

$S=\frac{1}{1-x+x^2}$ is coefficient of $x^n$

=coefficient of $x^n$ in $\frac{1+x}{1+x^3}$

= coefficient of $x^n$ in $(1+x){(1+x^3)}^{-1}$ and since n is a multiple of $3$, the co efficient of $x^n$ is unity and is negative when n is odd, and positive when n is even.

$S={(-1)}^n$