My attempt at the solution is to let P(n) be $10^{3n} + 13^{n+1}$
P(1)= $10^3 + 13^2 = 1169$
Thus P(1) is true.
Suppose P(k) is true for all $k \in N$ $\Rightarrow P(k) = 10^{3k} + 13^{k+1} = 10^{3k} + 13\cdot13^{k}$
$P(k+1) = 10^{3k+3} + 13^{k+2} \\ P(k+1) = 1000\cdot10^{3k} + 169\cdot13^k \\ P(k+1) = (10^{3k} + 13\cdot13^k) + 999\cdot10^{3k} + 12\cdot13^{k+1}$
My solution is not divisible by 7. I've always use this method to solve these types of questions. Can someone point out my error?
On
It would be convenient if we could find $a,b,c \in \mathbb Z$ such that: \begin{align*} 1000 &= a + 7b \\ 169 &= 13a + 7c \end{align*} If such integers existed, then we'd be done, since: $$ P(k + 1) = 1000 \cdot 10^{3k} + 169 \cdot 13^k = a(\underbrace{10^{3k} + 13 \cdot 13^{k}}_{\text{divisible by $7$}}) + 7(b \cdot 10^{3k} + c \cdot 13^k) $$ Indeed, by modding each term by $7$ in the first two equations, we find in both cases that $a = 6$, which yields $b = 142$ and $c = 13$, as desired.
HINT:
$$P(k+1)-13P(k)=10^{3(k+1)}+13^{k+1+1}-13[10^{3k}+13^{k+1}]$$
$$=10^{3k}[1000-13]$$
Now $1000-13$ is divisible by $7$
So, $P(k+1)$ will be divisible by $7\iff P(k)$ is