Question

Use the principle of induction to prove that$1 + 3 + 5 + ... + (2n - 1) = n^{2}$ for all n $\in \mathbb{N}$.​

Answer 1

Step-by-step explanation:

Given: 1 + 3 + 5 + ..... + (2n - 1) = n².

Let the given statement be P(n). Then,

P(n) = 1 + 3 + 5 + ... + (2n - 1) = n²

Substitute n = 1 in the given statement, we get

LHS: [2(1) - 1)] = 1

RHS: (1)² = 1

So, the statement holds for n = 1.

Let P(k) be true.

P(k) = 1 + 3 + 5 + .... + (2k - 1) = k²

For P(k + 1),

LHS = 1 + 3 + 5 + ... + (2k - 1) + (2(k + 1) - 1)

       = k² + (2k + 2 - 1)

       = k² + 2k + 1

       = (k + 1)²

∴ P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of induction, P(n) is true for all n ∈ N.

Hope it helps!

Answer 2

Answer:

ʜᴇʏᴀ

Step-by-step explanation:

Given: 1 + 3 + 5 + ..... + (2n - 1) = n².

Let the given statement be P(n). Then,

P(n) = 1 + 3 + 5 + ... + (2n - 1) = n²

Substitute n = 1 in the given statement, we get

LHS: [2(1) - 1)] = 1

RHS: (1)² = 1

So, the statement holds for n = 1.

Let P(k) be true.

P(k) = 1 + 3 + 5 + .... + (2k - 1) = k²

For P(k + 1),

LHS = 1 + 3 + 5 + ... + (2k - 1) + (2(k + 1) - 1)

      = k² + (2k + 2 - 1)

      = k² + 2k + 1

      = (k + 1)²

∴ P(k + 1) is true, whenever P(k) is true.

Hence, by the principle of induction, P(n) is true for all n ∈ N.