Prove this mathematical induction?

Question

$$ \sum_{j=1}^{n} 2^j = 2 \left( 2^{n} - 1 \right) $$ Could someone explain how to prove this problem fully?

Link to problem image

Answer 1

Use the relations $$ 1 - q^n = (1 - q)\left( 1 + q + \cdots + q^{n-1}\right) $$ and $$ \sum_{j=1}^{n} 2^j = 2 \left( \sum_{j=0}^{n-1} 2^{j} \right) = 2 \frac{1 - 2^{n}}{1 - 2} = 2 \left( 2^{n} - 1 \right) $$

Answer 2

When $n=1$,

$$L.H.S.=\sum^{(1)}_{j=1}2^j=2=2\big(2^{(1)}-1\big)=R.H.S.$$ $\therefore$The proposition is true for $n=1$.

Assume that the proposition is true when $n=k\in\mathbb{Z^+},$ i.e. $$\sum^k_{j=1}2^j=2\big(2^k-1\big).$$ When $n=k+1$,

$\begin{split}L.H.S.&=\sum^{(k+1)}_{j=1}2^j\\&=2\big(2^k-1\big)+2^{k+1}\\&=2\big(2^{(k+1)}-1\big)\\&=R.H.S.\end{split}$

$\therefore$The proposition is true for $n=k+1$ when the proposition is true for $n=k$.

$\therefore$The proposition is true by the principle of mathematical innduction.