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

S_n = 3 + 7 + 13 + 21 + 31 + … + a_{n - 1} + a_n …(1)

S_n = 0 + 3 + 7 + 13 + 21 + 31 + … + a_{n - 2} + a_{n - 1} + a_n …(2)

Subtracting (2) from (1)

S_n - S_n = (3 - 0) + [(7 - 3) + (13 - 7) + (21 - 13) + … + (a_{n - 1} - a_{n - 2})

+ (a_n - a_{n - 1}) - a_n

⇒ 0 = 3 + [4 + 6 + 8 + … a_{n - 1}] - a_n

⇒ a_n = 3 + [4 + 6 + 8 + … a_{n - 1}] …(3)

Now, 4 + 6 + 8 + … a_{n - 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)

∴

a_n = 3 + [4 + 6 + 8 + … a_{n - 1}]

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

= 3 + n² + 2n - n - 2

= 3 + n² + n - 2

= n² + n + 1

Now,

S_n

Thus, the required sum is .