Answer:
To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;
3·4ⁿ is divisible by 3 and 51 is divisible by 3
Where we have;
[tex]S_{(n)}[/tex] = 3·4ⁿ + 51
[tex]S_{(n+1)}[/tex] = 3·4ⁿ⁺¹ + 51
[tex]S_{(n+1)}[/tex] - [tex]S_{(n)}[/tex] = 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ
[tex]S_{(n+1)}[/tex] - [tex]S_{(n)}[/tex] = 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ
∴ [tex]S_{(n+1)}[/tex] - [tex]S_{(n)}[/tex] is divisible by 9
Given that we have for S₀ = 3×4⁰ + 51 = 63 = 9×7
∴ S₀ is divisible by 9
Since [tex]S_{(n+1)}[/tex] - [tex]S_{(n)}[/tex] is divisible by 9, we have;
[tex]S_{(0+1)}[/tex] - [tex]S_{(0)}[/tex] = [tex]S_{(1)}[/tex] - [tex]S_{(0)}[/tex] is divisible by 9
Therefore [tex]S_{(1)}[/tex] is divisible by 9 and [tex]S_{(n)}[/tex] is divisible by 9 for all positive integers n
Step-by-step explanation:
