Show with that induction $\sum_{k=0}^{n} 2{^{2k}}=\frac{2^{2n+2}-1}{3}$, for n = 0,1,2... Show with that induction that 
$\sum_{k=0}^{n} 2{^{2k}}=\frac{2^{2n+2}-1}{3}$, for n = 0,1,2...
My attempt
n = 0 
$LHS = n^0 = 1$
$RHS = (2^2 - 1 ) / 3  = 1$
n = p 
$LHS_{p} = 2^{2(0)}+2^{2(1)}+2^{2(2)}+...+2^{2(p)}$
$RHS_{p} = \frac{2^{2p+2}-1}{3}$
n = p + 1
$LHS_{p+1} = 2^{2(0)}+2^{2(1)}+2^{2(2)}+...+2^{2(p)++2^{2(p+1)}$
$RHS_{p+1} = \frac{2^{2(p+1)+2}-1}{3} $
Now i need to show that:
$RHS_{p+1} = RHS_{p}+2^{2(p+1)}$
$RHS_{p+1} = \frac{2^{2p+2}-1}{3} +2^{2(p+1)}$
Anyone see how i can rewrite?
$\frac{2^{2(p+1)+2}-1}{3} $ to be equal to 
$\frac{2^{2p+2}-1}{3} +2^{2(p+1)}$

$$\frac{2^{2(p+1)+2}-1}{3}=\frac{4(2^{2(p+1)})-1}{3}=\frac{2^{2(p+1)}-1+3(2^{2(p+1)})}{3}=\frac{2^{2(p+1)}-1}{3}+2^{2(p+1)}$$
