Question

Find the sum to $n$ terms of the sequence $(x+1/x{)}^2,(x^2+1/x^2{)}^2,(x^2+1/x^3{)}^2,......$

Answer 1

Let $S_n$ denote the sum to $n$ terms of the given sequence.
Then,
$S_n={(x+\frac{1}{x})}^2+{(x^2+\frac{1}{x^2})}^2+{(x^3+\frac{1}{x^3})}^2+....+{(x^n+\frac{1}{x^n})}^2$
$=(x^2+\frac{1}{x^2}+2)+(x^4+\frac{1}{x^4}+2)+(x^6+\frac{1}{x^6}+2)+.....+(x^{2n}+\frac{1}{x^{2n}}+2)$
$=(x^2+x^4+x^6+.....+x^{2n})+(\frac{1}{x^2}+\frac{1}{x^4}+\frac{1}{x^6}+.....+\frac{1}{x^{2n}})+\underset{ntimes}{(2+2+....+n)}$
$=x^2\left(\frac{(x^2{)}^n-1}{x^2-1}\right)+\frac{1}{x^2}\left(\frac{(1/x^2{)}^n-1}{\left(1/x^2\right)-1}\right)+2n$
$=x^2\left(\frac{x^{2n}-1}{x^2-1}\right)+\frac{1}{x^{2n}}\left(\frac{1-x^{2n}}{1-x^2}\right)+2n$
$=x^2\left(\frac{x^{2n}-1}{x^2-1}\right)+\frac{1}{x^{2n}}\left(\frac{x^{2n}-1}{x^2-1}\right)+2n$
$=\left(\frac{x^{2n}-1}{x^2-1}\right)(x^2+\frac{1}{x^{2n}})+2n$.