Question

Show that the sum of the products $n-r$ together of the $n$ quantities $a,a^2,a^3,\dots a^n$ is
$\frac{(a^{r+1}-1)(a^{r+2}-1)\dots (a^n-1)}{(a-1)(a^2-1)\dots (a^{n-r}-1)}{a}^{\frac{1}{2}(n-r)(n-r+1)}$.

Answer 1

To Show that the sum of the products n - r together of the n quantities $a,a^2,a^3...........a^n$ is

$\frac{(a^{r+1}-1)(a^{r+2}-1)......(a^n-1)}{(a-1)(a^2-1).......(a^{n-r}-1)}{a}^{\frac{1}{2}(n-r)(n-r+1)}$

The Sum of the Products is the Coefficient of $x^r$ in the Expansion of

$(x+a)(x+a^2)......(x+a^n)=x^n+A_1{x}^{n-1}+.....+A_{n-r-1}{x}^{r+1}+A_{n-r}{x}^r$

Write $\frac{x}{a}$ for x , then Since

$\frac{x}{a}+a^r=\frac{1}{a}(x+a^{r+1})$ , we have

$\frac{1}{a^n}(x+a^2)(x+a^3).....(x+a^{n+1})={\left(\frac{x}{a}\right)}^n+A_1{\left(\frac{x}{a}\right)}^{n-1}+A_2{\left(\frac{x}{a}\right)}^{n-2}......$

$\therefore$

$(x+a^2)(x+a^3).......(x+a^{n+1})=x^n+A_{1}a{x}^{n-1}+.....+A_{n-r-1}{a}^{n-r-1}{x}^{r+1}+....$

$\therefore$

$(x+a)(x^n+A_{1}a{x}^{n-1}+.....A_{n-r-1}{a}^{n-r-1}{x}^{r+1}+A_{n-r}{a}^{n-r}{x}^r+....)$

$(x+a^{n+1})(x^n+A_1{x}^{n-1}+.....A_{n-r-1}{x}^{r+1}+A_{n-r}{x}^r+....)$

On Equating the Coefficients of $x^{r+1}$; Then ,

$A_{n-r}{a}^{n-r}+A_{n-r-1}{a}^{n-r}=A_{n-r-1}{a}^{n+1}+A_{n-r}$

$\therefore$

$A_{n-r}(a^{n-r}-1)=A_{n-r-1}{a}^{n-r}(a^{r+1}-1)$

$A_{n-r}=A_{n-r-1}{a}^{n-r}\frac{(a^{r+1}-1)}{(a^{n-r}-1)}$

On Putting r + 1 in Place of r , Then ;

$A_{n-r-1}=A_{n-r-2}{a}^{n-r-1}\frac{(a^{r+2}-1)}{(a^{n-r-1}-1)}$

..................... = ............................................................................................................

$A_2=A_1{a}^2\left(\frac{a^{n-1}-1}{a^2-1}\right)$

$A_1=a\left(\frac{a^n-1}{a-1}\right)$ , Since $A_0=1$

Now on Multiplying these Equations Together , we have

$A_{n-r}.A_{n-r-1}.......A_1=\frac{(a^{r+1}-1)(a^{r+2}-1).....(a^n-1)}{(a-1)(a^2-1).....(a^{n-r}-1)}{a}^{\frac{1}{2}(n-r)(n-r+1)}$