Proving that $\sum_{j=0}^n 2^j=2^{n+1}-1$ for $n\geq 0$ by induction

Question

Solving $S=$ $\displaystyle \sum_{j=0}^n$ $2$$^j$ = $2$$^n$$^+$$^1$ $-1$

So I was able to find the basis and the the RHS but I'm not sure how I should go about solving the LHS. Since I have K+1 in the sigma

Solving $2$$^K$$^+$$^1$ $-1$ + $\displaystyle \sum_{j=0}^{n+1}2^j$ = $2$$^K$$^+$$^2$ $-1$

Answer 1

See if you can follow the meat of the induction proof below (getting from the LHS to the RHS): \begin{align} \sum_{j=0}^{k+1}2^j &= \sum_{j=0}^k 2^j+2^{k+1}\tag{by defn. of $\Sigma$}\\[0.5em] &= (2^{k+1}-1)+2^{k+1}\tag{by inductive hypothesis}\\[0.5em] &= 2\cdot2^{k+1}-1\tag{group like terms}\\[0.5em] &= 2^{k+2}-1\tag{exponent law}\\[0.5em] &= 2^{(k+1)+1}-1.\tag{simplify} \end{align}