Question

if a1=1,a2=1, and a3=9 for the sequence defined by an+3=-an+2 +2an+1+8an, then prove that for every value of n,an is a perfect square

Answer 1

Please write correct question for getting quickly answer .
Your question is ---------> If a1=1,a2=1, and a3=9 for the sequence defined by an+3=-an+2 +2an+1+8an, then prove that for every value of n,an+3 is a perfect square?

Given, a₁ = 1 , a₂ = 1 and a₃ = 9
expression is $\bold{a_{n+3}=-a_{n+2}+2a_{n+1}+8a_n}$
Putting n = 1
a₄ = -a₃ + 2a₂ + 8a₁
= -9 + 2 × 1 + 8 × 1
= -9 + 2 + 8
= -9 + 10
= 1
a₄ = 1²

Putting n = 2
a₅ = -a₄ + 2a₃ + 8a₂
= -1 + 2 × 9 + 8 × 1
= -1 + 18 + 8
= 25
a₅ = 5²

Putting n = 3
a₆ = -a₅ + 2a₄ + 8a₃
= - 25 + 2 × 1 + 8 × 9
= -25 + 2 + 72
= 49
a₆ = 7²

Here we observed for every integer value of n , $\bold{a_{n+3}}$ is a perfect square