Prove that displaystylesum_{j=1}^{n} 2^{j}=2^{n+1}-2 forallgeq1

Step 1
To proof: ${\displaystyle \sum _{j=1}^n{2}^j=2^{n+1}-2}$ for all positive integers n.
Proof by induction
Let P(n) be the statement "${\displaystyle \sum _{j=1}^n{2}^j=2^{n+1}-2}$".
Basis step n=1
${\displaystyle \sum _{j=1}^n{2}^j={\displaystyle \sum _{j=1}^1{2}^j=2^1=2}}$
$2^{n+1}-2=2^{1+1}-2=2^2-2=2$
Thus P(1) is true.
Inductive step
Let P(k) be true
${\displaystyle \sum _{j=1}^k{2}^j=2^{k+1}-2}$
We need to proof that P(k+1) is true
${\displaystyle \sum _{j=1}^{k+1}{2}^j}$
$=[{\displaystyle \sum _{j=1}^{k+1}{2}^j]+2^{k+1}}$
$=2^{k+1}-2+2^{k+1}\text{Since P(k) is true}$
$=2\times 2^{k+1}-2\text{Combine loke terms}$
$=2^{k+1+1}-2$
$=2^{(k+1)+1}-2$
Thus P(k+1) is true
By the principle if mathematical induction, P(n) is true for all positive integers n.