Question

The first term of an ap of consecutive integers is p^2 + 1. then find the sum of 2p + 1 terms of ts series

Answer 1

The formula for finding the sum to the nth term of an AP is


$Sn = \frac{n}{2} [2a+(n-1)d]$


In this question:

a = p² + 1

d = 1 (They are consecutive integers)

n = 2p + 1



Substitute these in the above formula


Sn = (2p+1)/2{2(p² + 1) + (2p+1 -1)(1)}


Sn = (2p+1)/2{2p² + 2 + 2p}


Sn = (2p+1)(p²+p+1)


Sn = 2p³ + 3p² + 1

Answer 2

Step-by-step explanation:

The formula for finding the sum to the nth term of an AP is

Sn = \frac{n}{2} [2a+(n-1)d]Sn=

2

n

[2a+(n−1)d]

In this question:

a = p² + 1

d = 1 (They are consecutive integers)

n = 2p + 1

Substitute these in the above formula

Sn = (2p+1)/2{2(p² + 1) + (2p+1 -1)(1)}

Sn = (2p+1)/2{2p² + 2 + 2p}

Sn = (2p+1)(p²+p+1)

sn= p³+(p+1)³