Number pattern and sequence

Question

The first four terms in a sequence of numbers are given below.

P1=0^2+4 = 4

P2 = 1^2 +7=8

P3= 2^2 +10 = 14

P4 = 3^2 +13= 22

a) Explain why the value of Pn will never be an odd number for all values of n.

Answer 1

Pn = (n-1)^2 +4 + 3(n-1) = n^2 -2n + 1 + 4 + 3n -3 = n^2 +n +2

Pn = n^2 +n + 2 = n(n+1) + 2

the nth term = n(n+1) + 2

P1 =4

P2 =4+2+2 = 8

P3 = 9+3+2 = 14

P4 = 16+4+2 = 22

if n is even n(n+1) is even. n+1 is odd. but an even times an odd is always even. Add 2 and it's still even

if n is odd then n(n+1) is even. n+1 is even and an odd times an even is always even. Add 2 & it's still even