Find the sum of the first n odd natural numbers. Hence find 1 + 3 + 5 + ... + 101.

5. Find the sum of the first n odd natural numbers.Hence find 1 + 3 + 5 + ... + 101.

Sol. The first n odd natural numbers are as follows: 1, 3, 5, 7, ............., n

Here, a = 1, d = t2 – t1 = 3 – 1 = 2

Sn = n/2[2a + (n – 1)d]

∴  Sn = n/2 [2 (1) + (n – 1) 2]

∴  Sn = n/2[2 + 2n – 2]

∴ Sn = n/2 [2n]

∴ Sn = n² ......(i)

Now, we have,

1 + 3 + 5 + ........ + 101

Let, 101 be the nth term of A.P.

∴ tn = 101

tn = a + (n – 1) d

∴  101 = a + (n – 1) d

∴  101 = 1 + (n – 1) 2

∴  101 = 1 + 2n – 2

∴  101 = 2n – 1

∴  101 + 1 = 2n

∴  2n = 102

∴  n = 51

∴ 101 is the 51st term of A.P.,

Now,  We have to find sum of 51 terms i.e. S51,

Sn = n² [From (i)]

∴  S51 = (51)²

∴  S51 = 2601