Question

Show that for any non-zero rational number $x$, the following relation holds true:
$1+\frac{x}{2}+\frac{x(x-1)}{2.4}+\frac{x(x-1)(x-2)}{\mathrm{2.4.6}}+....=1+\frac{x}{3}+\frac{x(x+1)}{3.6}+\frac{x(x+1)(x+2)}{\mathrm{3.6.9}}+....$

Answer 1

$1+\frac{x}{2}+\frac{x(x-1)}{2.4}+\frac{x(x-1)(x-2)}{\mathrm{2.4.6}}+......$

$=1+x.\frac{1}{2}+\frac{x(x-1)}{1.2}\frac{1}{2^2}+\frac{x(x-1)(x-2)}{\mathrm{1.2.3}}\frac{1}{2^3}+......$

$={(1+\frac{1}{2})}^x$ [Since $x$ is a rational number]

$={\left(\frac{3}{2}\right)}^x$

$={\left(\frac{2}{3}\right)}^{-x}$

$={(1-\frac{1}{3})}^{-x}$ [$-x$ is also rational number.]

$=1+x.\frac{1}{3}+\frac{x(x+1)}{1.2}\frac{1}{3^2}+\frac{x(x+1)(x+2)}{\mathrm{1.2.3}}\frac{1}{3^3}+......$

$=1+\frac{x}{3}+\frac{x(x+1)}{3.6}+\frac{x(x+1)(x+2)}{\mathrm{3.6.9}}+......$