So we have been given the discrete time signal and we have to compute the energy
$y[n]=(\frac 1 2 )^n u[n] -(-\frac 1 2)^{n-1}u[n-1]$
The answer is supposed to be $\frac {28} {15}$ but i cant seem to go there
From what i know since $\sum _0 ^{N-1} r^k$ if $r < 1 $ = $ \frac 1 {1-r}$ And since E=$\sum x^2[n]$ then we have
E=$\sum x^2[n]= (\frac 1 2 )^{2n} + 2(\frac 1 2 )^n(-\frac 1 2 )^n + (-\frac 1 2 )^{2n} (\frac 1{(\frac 1 2 )^2})$
And from here i thought we just apply the formulas and thats about it but i cant seem to do it
We have $$ y^2[n]=\left(\frac12\right)^{2n}u[n]+\left(-\frac12\right)^{2n-2}u[n-1]-2\left(\frac12\right)^{n}\left(-\frac12\right)^{n-1}u[n]u[n-1] $$ so that $$ \begin{align*} E&=\sum_{n=-\infty}^\infty y^2[n]\\ &=\sum_{n=0}^\infty\left(\frac14\right)^{n}+\sum_{n=1}^\infty\left(\frac14\right)^{n-1}-2\sum_{n=1}^\infty\left(\frac12\right)^{n}\left(-\frac12\right)^{n-1}\\ &=\sum_{n=0}^\infty\left(\frac14\right)^{n}+\color{red}{\sum_{n=0}^\infty\left(\frac14\right)^{n}}-2\sum_{n=1}^\infty\left(\frac12\right)^{n}\left(\frac12\right)^{n-1}(-1)^{n-1}\qquad (\color{red}{n-1=i,\,i\leftrightarrow n})\\ &=2\sum_{n=0}^\infty\left(\frac14\right)^{n}-\require{cancel}\cancel{2}\color{blue}{\sum_{n=0}^\infty\left(\frac12\right)^{n+\cancel{1}}\left(\frac12\right)^{n}(-1)^{n}}\qquad\qquad\qquad\quad(\color{blue}{n-1=i,\,i\leftrightarrow n})\\ &=2\sum_{n=0}^\infty\left(\frac14\right)^{n}-\sum_{n=0}^\infty\left(-\frac14\right)^{n}\\ &=2\cdot\frac43-\frac45=\boxed{\frac{28}{15}} \end{align*} $$ using $\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x}$, for $|x|<1$.