Q. 155.0( 3 Votes )

# 1 + 2 + 2^{2} + … 2^{n} = 2^{n+1} – 1 for all natural numbers n.

Answer :

Given; P(n) is 1 + 2 + 2^{2} + … 2^{n} = 2^{n+1} – 1.

P(0) = 1 = 2^{0+1} − 1 ; it’s true.

P(1) = 1 + 2 = 3 = 2^{1+1} − 1 ; it’s true.

P(2) = 1 + 2 + 2^{2} = 7 = 2^{2+1} − 1 ; it’s true.

P(3) = 1 + 2 + 2^{2} + 2^{3} = 15 = 2^{3+1} − 1 ; it’s true.

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

⇒ P(k+1) is 1 + 2 + 2^{2} + … 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 + 2^{2} + … 2^{n} = 2^{n+1} – 1 is true for all natural numbers n.

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