prove that $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 Prove that a number $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 for every natural $n\ge2$. I am not sure how to start.

The base case is true, $T_0=4+2+1+64-8=63$ is a multiple of $7$.
Now let us relate $T_{n+1}$ to $T_n$:
$$T_{n+1}=2.2^n+2.2^{n-1}+2.2^{n-2}+8.8^n-8.8^{n-2}.$$
Subtracting $T_n$, we have
$$T_{n+1}-T_n=2^n+2^{n-1}+2^{n-2}+7.8^n-7.8^{n-2}.$$
We can discard the last two terms, obviously multiples of $7$. Remains to prove that $2^n+2^{n-1}+2^{n-2}$ is a multiple of $7$.
The base case is true, $U_0=4+2+1=7$.
Now let us relate $U_{n+1}$ to $U_n$:
$$U_{n+1}=2.2^n+2.2^{n-1}+2.2^{n-2}=2.U_n.$$
As $U_n$ is a multiple of $7$, so is $U_{n+1}$.
