# Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

This given series is neither an AP nor GP.

Let

Sn = 3 + 7 + 13 + 21 + 31 + … + an - 1 + an …(1)

Sn = 0 + 3 + 7 + 13 + 21 + 31 + … + an - 2 + an - 1 + an …(2)

Subtracting (2) from (1)

Sn - Sn = (3 - 0) + [(7 - 3) + (13 - 7) + (21 - 13) + … + (an - 1 - an - 2)

+ (an - an - 1) - an

0 = 3 + [4 + 6 + 8 + … an - 1] - an

an = 3 + [4 + 6 + 8 + … an - 1] …(3)

Now, 4 + 6 + 8 + … an - 1 is an AP.

Where,

first term(a) = 4

common difference(d) = 6 - 4 = 2

We know that,

Sum of n terms of AP = (n/2)[2a + (n - 1)d]

putting n = n - 1, a = 4, d =2

[4 + 6 + 8 + … to (n - 1) terms] = (n - 1)/2 × [2a + (n - 1 - 1)d]

= (n - 1)/2 × [2(4) + (n - 2)2]

= (n - 1)/2 × [8 + 2n - 4]

= (n - 1)/2 × [2n + 4]

= (n - 1)/2 × 2(n + 2)

= (n - 1)(n + 2)

an = 3 + [4 + 6 + 8 + … an - 1]

= 3 + (n - 1)(n + 2)

= 3 + n2 + 2n - n - 2

= 3 + n2 + n - 2

= n2 + n + 1

Now,

Sn

Thus, the required sum is .

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