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

\(\displaystyle\sum_{j=1}^{k+1} 2^{j}\)

\(=[\displaystyle\sum_{j=1}^{k+1} 2^{j}]+2^{k+1}\)

\(=2^{k+1}-2+2^{k+1} \ \text{Since P(k) is true}\)

\(=2\times2^{k+1}-2 \ \text{Combine loke terms}\)

\(=2^{k+1+1}-2\)

\(=2^{(k+1)+1}-2\)

Thus P(k+1) is true

By the principle if mathematical induction, P(n) is true for all positive integers n.

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

\(\displaystyle\sum_{j=1}^{k+1} 2^{j}\)

\(=[\displaystyle\sum_{j=1}^{k+1} 2^{j}]+2^{k+1}\)

\(=2^{k+1}-2+2^{k+1} \ \text{Since P(k) is true}\)

\(=2\times2^{k+1}-2 \ \text{Combine loke terms}\)

\(=2^{k+1+1}-2\)

\(=2^{(k+1)+1}-2\)

Thus P(k+1) is true

By the principle if mathematical induction, P(n) is true for all positive integers n.