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Question -

1 + 3 + 5 + … + (2n – 1) = n2 i.e.,the sum of first n odd natural numbers is n2.



Answer -

Let P (n): 1 + 3 + 5 + … + (2n – 1) = n2

Let us check P (n) is true for n = 1

P (1) = 1 =12

1 = 1

P (n) is true for n = 1

Now, Let’s check P (n) is true for n = k

P (k) = 1 + 3 + 5 + … + (2k – 1) = k2 … (i)

We have to show that

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1 = (k + 1)2

Now,

1 + 3 + 5 + … + (2k – 1) + 2(k + 1) – 1

= k2 + (2k + 1)

= k2 + 2k + 1

= (k + 1)2

P (n) is true for n = k + 1

Hence, P (n) is true for all n N.

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