# 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 + … 2k = 2k+1 – 1 is true;

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.

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

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