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

Given; P(n) is 1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1.

P(0) = 1 = 2⁰⁺¹ − 1 ; it’s true.

P(1) = 1 + 2 = 3 = 2¹⁺¹ − 1 ; it’s true.

P(2) = 1 + 2 + 2² = 7 = 2²⁺¹ − 1 ; it’s true.

P(3) = 1 + 2 + 2² + 2³ = 15 = 2³⁺¹ − 1 ; it’s true.

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

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

= 2×2^k+1 – 1

= 2^(k+1)+1 – 1

⇒ P(k+1) is true when P(k) is true.

∴ By Mathematical Induction 1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1 is true for all natural numbers n.