Show that $6^{(k+1)+1} + 7^{2(k+1)-1}$ is a multiple of $43$

Question

Show that $6^{(k+1)+1} + 7^{2(k+1)-1}$ is a multiple of $43$ knowing that $6^{k+1} + 7^{2k-1}$ is a multiple of 43.

I have written it as $6^{k+1} 6^1 + 7^{2k-1} 7^2$ but idk what to do after.

Answer 1

\begin{align*} \left(6^{(k+1)+1} + 7^{2(k+1)-1}\right) - \left(6^{k+1} + 7^{2k-1}\right) &= 6^{k+1}(6-1) + 7^{2k-1}(49-1) \\ &= 5(6^{k+1}+7^{2k-1}) + 43 \cdot 7^{2k-1} \end{align*}