The first term of an A.P of consecutive integers is p² + 1. The sum of (2p + 1) terms of this series can be expressed as

(1) (P + 1)²

(2) (p + 1)³

(3) (2p + 1)(p + 1)²

(4) p³ + (p + 1)³

Solution:

Given first term of AP, a = p²+1

Here d = 1, since integers are consecutive.

Sum of first n terms of AP, S_n = (n/2)(2a+(n-1)d)

S_2p+1 = ((2p+1)/2)(2a+(2p+1-1)d)

= ((2p+1)/2)(2(p²+1)+2p)

= (2p+1)(p²+1+p)

= 2p³+p²+2p+1+2p²+p

= p³ + p³ + 3p²+ 3p +1

= p³ + (p+1)³

Hence option (4) is the answer.