1 + 2 + 22 + … 2n = 2n+1 – 1

According to the question,

P(n) is 1 + 2 + 22 + … 2n = 2n+1 – 1.

So, substituting different values for n, we get,

P(0) = 1 = 20+1 − 1 Which is true.

P(1) = 1 + 2 = 3 = 21+1 − 1 Which is true.

P(2) = 1 + 2 + 22 = 7 = 22+1 − 1 Which is true.

P(3) = 1 + 2 + 22 + 23 = 15 = 23+1 − 1 Which is true.

Let P(k) = 1 + 2 + 22 + … 2k = 2k+1 – 1 be true;

So, we get,

⇒ P(k+1) is 1 + 2 + 22 + … 2k + 2k+1 = 2k+1 – 1 + 2k+1

= 2×2k+1 – 1

= 2(k+1)+1 – 1

⇒ P(k+1) is true when P(k) is true.

Therefore, by Mathematical Induction,

1 + 2 + 22 + … 2n = 2n+1 – 1 is true for all natural numbers n.