Can you help to prove this equation by induction

Question

Can you help to prove by induction that $(9^n - 1)(n^3 + 3n^2+ 2n + 6)$ is divisible by $48$ .

I know that I can assume that it holds for $n$, and try to prove than it holds for $(n+1)$. That is where I got stuck.

Answer 1

let $T(n)=(9^n-1)(n^3+3n^2+2n+6)$ and $$48|T(n)$$ we have to show that $48|T(n+1)$ now compute $$T(n+1)-T(n)$$ and this is $$(9^n-1)\cdot 3(n+1)(2+n)+8\cdot 9^n(n+4)(n^2+2n+3)$$

Answer 2

$$9^{n+1}-1=9\cdot9^n-1=8\cdot9^n+9^n-2\equiv9^n-1\mod 8.$$

and

$$(n+1)^3 + 3(n+1)^2+ 2(n+1) + 6=n^3+6n^2+11n+12\\ =3n^2+9n+6+n^3+3n^2+2n+6\\ =3(n+1)(n+2)+n^3+3n^2+2n+6. $$

Then one of $n+1$, $n+2$ is even and

$$(n+1)^3 + 3(n+1)^2+ 2(n+1) + 6=n^3+3n^2+2n+6\mod 6.$$

Finally, the base cases show that $9^n-1$ is a multiple of $8$ and $n^3+3n^2+2n+6$ a multiple of $6$.