Solving/Proving Recurrence Relations Given $a_{k-1}=7*5^{k-1}-3*8^{k-1}$ and
$a_k=7*5^k-3*8^k$
for the recurrence $a_n=13a_{n-1}-40a_{n-2}$
prove that the next term is $a_{k+1}=7*5^{k+1}-3*8^{k+1}$
I am kind of stuck right now. This is what I have so far. If anyone could give me advice on where I should go from here. Thanks
$a_{k+1}=13(7*5^k-3*8^k)-40(7*5^{k-1}-3*8^{k-1})$
$=13*7*5^k-13*3*8^k-40*7*5^{k-1}+40*3*8^{k-1}$
$=7(13*5^k-40*5^{k-1})+3(-13*8^k+40*8^{k-1})$
$=7(13*5-40)*5^{k-1}+3(-13*8+40)*8^{k-1}$

Express $a_{k+1}$ as a linear combination of $5^{k-1}$ and $3^{k-1}$, rework the coefficients and extract the powers of $5$ and $3$.
$$13\,(7\cdot5^k-3\cdot8^k)-40\,(7\cdot5^{k-1}-3\cdot8^{k-1})
\\=(13\cdot7\cdot5-40\cdot7)\,5^{k-1}-(13\cdot3\cdot8-40\cdot3)\,3^{k-1}
\\=175\cdot5^{k-1}-192\cdot 3^{k-1}
\\=7\cdot5^2\cdot5^{k-1}-3\cdot3^2\cdot8^{k-1}
\\=7\cdot5^{k+1}-3\cdot8^{k+1}$$
