Question

Find the sum of ${(x+\frac{1}{x})}^3+{(x^2+\frac{1}{x^2})}^3+...+{(x^n+\frac{1}{x^n})}^3$

• A. $\frac{1}{(1-x^2)}[x^3-x^{3(n+1)}-1+\frac{1}{x^{2n}}]+\frac{3}{(1-x)}[x-x^{n+1}-1+\frac{1}{x^n}]$
• B. $\frac{1}{(1-x^3)}[x^3-x^{3(n+1)}-1+\frac{1}{x^{3n}}]+\frac{3}{(1-x)}[x-x^{n+1}-1+\frac{1}{x^n}]$
• C. $\frac{1}{(1-x^{})}[x^2-x^{2(n+1)}-1+\frac{1}{x^{2n}}]+\frac{3}{(1-x)}[x-x^{n+1}-1+\frac{1}{x^n}]$
• D. $\frac{1}{(1-x^2)}[x^2-x^{2(n+1)}-1+\frac{1}{x^{3n}}]+\frac{3}{(1-x)}[x-x^{n+1}-1+\frac{1}{x^n}]$

Answer 1

The correct option is B $\frac{1}{(1-x^3)}[x^3-x^{3(n+1)}-1+\frac{1}{x^{3n}}]+\frac{3}{(1-x)}[x-x^{n+1}-1+\frac{1}{x^n}]$

$S$ $=$ ${(x+\frac{1}{x})}^3+{(x^2+\frac{1}{x^2})}^3+...+{(x^n+\frac{1}{x^n})}^3$
Expanding the brackets and regrouping into $4$ series gives
$S$ =$(x^3+x^6+\cdots +x^{3n})+(\frac{1}{x^3}+\frac{1}{x^6}+\cdots +\frac{1}{x^{3n}})+3(x+x^2+\cdots +x^n)+3(\frac{1}{x}+\frac{1}{x^2}+\cdots +\frac{1}{x^n})$
$=x^3(1+x^3+x^6+\cdots +x^{3n-3})+\frac{1}{x^3}(1+\frac{1}{x^3}+\frac{1}{x^6}+\cdots +\frac{1}{x^{3n-3}})+3x(1+x+x^2+\cdots +x^{n-1})$
$+3\frac{1}{x}(1+\frac{1}{x}+\frac{1}{x^2}+\cdots +\frac{1}{x^{n-1}})$
$=\frac{x^3(1-x^{3n})}{1-x^3}+\frac{1}{x^3}\left(\frac{1-{\left(\frac{1}{x^3}\right)}^n}{1-\frac{1}{x^3}}\right)+3x\left(\frac{1-x^n}{1-x}\right)+\frac{3}{x}\left(\frac{1-\frac{1}{x^n}}{1-\frac{1}{x}}\right)$
$\therefore S=\frac{1}{(1-x^3)}[x^3-x^{3(n+1)}-1+\frac{1}{x^{3n}}]+\frac{3}{(1-x)}[x-x^{n+1}-1+\frac{1}{x^n}]$
Hence, option B.