Let $a_n=2^{2n}\left(2^{2n+1}-1 \right)$. Show induction of $n$
$$a_{2n+1}=256a_{2n-1}+60\left(16^n\right)$$
$$a_{2n+2}=256a_{2n}+240\left(16^n\right)$$
I tried $n=1$, $a_3=2^{2\cdot3}\left(2^{2\cdot3+1}-1\right)=2^6\cdot(2^7-1)=8128$ and $a_3=256a_1+60\cdot16^1=256\left(2^{2\cdot1}(2^{2\cdot 1+1}-1)\right)+960=256\cdot4\cdot7+960=8128$. This shows us that it is valid for $n=1$. And now? What is the hypothesis and the thesis?
You are given an equation for $a_n$ and two equations that it is claimed to satisfy. So plug $a_{2n-1}$ from the definition into the first and rearrange it to show that is $a_{2n+1}$. Your proof should start $$256a_{2n-1}+60(16^n)=256\cdot2^{2(2n-1)}\left(2^{2(2n-1)+1}-1\right)+60\cdot 16^n$$ and end with $=a_{2n+1}$. Then do it for $a_{2n+2}$You are not proving it by induction, because you are given the solution. You are trying to verify the solution.