Prove the statement by the Principle of Mathematical Induction :1 + 2

1.	Prove the statement by the Principle of Mathematical Induction :1 + 2 + 22 + … 2n = 2n+1 – 1 for all natural numbers n.
Answer» 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 + … 2^k = 2^k+1 – 1 be true; So, we get, ⇒ 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. Therefore, by Mathematical Induction, 1 + 2 + 2² + … 2ⁿ = 2ⁿ⁺¹ – 1 is true for all natural numbers n.

Prove the statement by the Principle of Mathematical Induction :1 + 2 + 22 + … 2n = 2n+1 – 1 for all natural numbers n.

Answer»

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 + … 2^k = 2^k+1 – 1 be true;

So, we get,

⇒ 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.

Therefore, by Mathematical Induction,

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