Calculation:

(2n + 1) + (2n + 3) + (2n + 5) + …… + (2n + 47) = 5280 

(2n + 5) – (2n + 3) = 2

(2n + 3) – (2n + 1) = 2

On the Left Hand Side, (2n + 1) + (2n + 3) + (2n + 5) + …… + (2n + 47) is an arithmetic progression with common difference 2 

First term of the arithmetic progression is (2n + 1) and the last term is (2n + 47)

Let the number of terms be T 

(2n + 47) = (2n + 1) + (T – 1)2

⇒ 2T – 2 = 2n + 47 – 2n – 1

⇒ 2T = 46 + 2

⇒ 2T = 48

⇒ T = 24

Number of terms of the arithmetic progression = 24

Sum of arithmetic progression = 24/2[(2n + 1) + (2n + 1) + (24 – 1)2]

⇒ 24/2[(2n + 1) + (2n + 1) + (24 – 1)2] = 5280

⇒ 24/2[4n + 2 + 23 × 2] = 5280

⇒ 24/2[4n + 48] = 5280

⇒ 48[n + 12] = 5280

⇒ n + 12 = 110

⇒ n = 110 – 12

⇒ n = 98

The value of 1 + 2 + 3 + …. + n =  1 + 2 + 3 + …. + 98

⇒ 98(98 + 1)/2

⇒ 49 × 99

⇒ 4851

∴ The value of 1 + 2 + 3 + …. + n is 4851

Formula Used:

nth term of an arithmetic progression = a + (n – 1)d

Here, n = Number of terms , a = First term, d = Common difference (Difference between two consecutive terms)

Sum of an arithmetic progression = n/2[a + a + (n – 1)d]

Sum of first n natural numbers = n(n + 1)/2