1 + 3 + 5 + ... + (2n – 1) = n2

1.	1 + 3 + 5 + ... + (2n – 1) = n2
Answer» Let the given statement P(n) be defined as P(n) : 1 + 3 + 5 +...+ (2n – 1) = n², for n ∈ N. Note that P(1) is true, since P(1) : 1 = 1² Assume that P(k) is true for some k ∈ N, i.e., P(k) : 1 + 3 + 5 + ... + (2k – 1) = k² Now, to prove that P(k + 1) is true, we have 1 + 3 + 5 + ... + (2k – 1) + (2k + 1) = k² + (2k + 1) = k² + 2k + 1 = (k + 1)² Thus, P(k + 1) is true, whenever P(k) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

Answer»

Let the given statement P(n) be defined as P(n) : 1 + 3 + 5 +...+ (2n – 1) = n², for n ∈ N. Note that P(1) is true, since

P(1) : 1 = 1²

Assume that P(k) is true for some k ∈ N, i.e.,

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

Now, to prove that P(k + 1) is true, we have

1 + 3 + 5 + ... + (2k – 1) + (2k + 1)

= k² + (2k + 1)

= k² + 2k + 1 = (k + 1)²

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

Hence, by the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

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