Question

If n is a multiple of $6$, show that each of the series $n-\frac{n(n-1)(n-2)}{\lfloor 3}\cdot 3+\frac{n(n-1)(n-2)(n-3)(n-4)}{\lfloor 5}\cdot 3^2-.....,$ $n-\frac{n(n-1)(n-2)}{\lfloor 3}\cdot \frac{1}{3}+\frac{n(n-1)(n-2)(n-3)(n-4)}{\lfloor 5}\cdot \frac{1}{3^2}-.....,$ is equal to zero.

Answer 1

$\Rightarrow$ Let ${(1+x)}^n=1+c_{1}x+c_2{x}^2+...$

${(1+ix)}^n=1+i{c}_{1}x-c_2{x}^2+i{c}_3{x}^3+c_4{x}^4+i{c}_5{x}^5-....$

${(1-ix)}^n=1-i{c}_{1}x-c_2{x}^2+i{c}_3{x}^3+c_4{x}^4-i{c}_5{x}^5-....$

$\therefore 2ix(c_1-c_3{x}^2+i{c}_5{x}^4-....)={(1+ix)}^n-{(1-ix)}^n$

Put $x^2=3$, so that $x=\sqrt{3}$, and let $S_1$ denote the value of the first series;also as usual

Let $w,w^2$ be the imaginary cube roots of unity;

so that $w=\frac{-1+\sqrt{-3}}{2};w^2=\frac{-1-\sqrt{-3}}{2}$

We have

$2i\sqrt{3}{S}_1={(1+\sqrt{-3})}^n-{(1-\sqrt{-3})}^n$

$={(-2w^2)}^n-{(-2w)}^n$

$=2^n-2^n=0$

when n is a multiple of $6$, for then

${(-w)}^n=1,{(-w^2)}^n=1$

Put $x^2=\frac{1}{3}$ and let $S_2$denote the sum of the series, them;-

$\frac{2i}{\sqrt{3}}{S}_2={(1+\frac{\sqrt{-1}}{\sqrt{3}})}^n-{(1-\frac{\sqrt{-1}}{\sqrt{3}})}^n$

$={\left(\frac{\sqrt{-3}-1}{\sqrt{-3}}\right)}^n-{\left(\frac{\sqrt{-3}+1}{\sqrt{-3}}\right)}^n$

$={\left(\frac{2w}{\sqrt{-3}}\right)}^n-{\left(\frac{-2w^2}{\sqrt{-3}}\right)}^n$

$=0$

if n is a multiple of $6$