Solve $$a_{n+2}-4a_{n+1}-5a_{n}=0,\quad\text{with $a_0=4$ and $a_1=2$}$$ and prove by induction that the formula found is equivalent to the given.
My calculations show that the particular solution is $a_n=3(-1)^{n}+5^{n}$. Now we need to prove it by induction.
For base case, let $n=0$ so $a_0=3+1=4$, and let $n=1$ so $a_1=-3+5=2$. Hence the base case is proven. Now let's see the inductive step:
What I have done:
Our hypothesis are: $a_h=3(-1)^h+5^h$ and $a_{h+1}=3(-1)^{h+1}+5^{h+1}$, and we want to show that $a_{h+2}=3(-1)^{h+2}+5^{5+2}$.
Proof If $a_{h+2}-4a_{h+1}-5a_{h}=0$ then \begin{align*}a_{h+2}&=4a_{h+1}+5a_{h}\\&=4(3(-1)^{h+1}+5^{h+1})+5(3(-1)^h+5^h)\\&=\color{red}{4\cdot3(-1)^{h+1}}+\color{blue}{4\cdot5^{h+1}}+\color{red}{5\cdot3(-1)^{h}}+\color{blue}{5^{h+1}}\\&=\color{red}{-4\cdot3(-1)^h+5\cdot3(-1)^h}\color{blue}{+5\cdot5^{h+1}}\\&=3(-1)^h+5^{h+2}\\&=3(-1)^2(-1)^h+5^{h+2}\\&=3(-1)^{h+2}+5^{h+2}.&\square\end{align*} However, my teammate did something quite different:
He says: $$a_{h+1}=4(5^h+3\cdot(-1)^h) + 5 (5^{h-1}+3\cdot(-1)^{h-1}).$$ I am not sure what he did. I think he assumed something with $a_{h-1}$ and $a_{h}$ but we need $a_{h}$ and $a_{h+1}$ as hypothesis, right?
How can we help him so he can achieve the correct proof?
Your teammate got $$a_{h+1}=4(5^h+3\cdot(-1)^h) + 5 (5^{h-1}+3\cdot(-1)^{h-1}) \\ =4 \cdot 5^h+12\cdot(-1)^h + 5 \cdot 5^{h-1}+15\cdot(-1)^{h-1}\\ =4 \cdot 5^h+12\cdot(-1)^h + 5^{h}-15\cdot(-1)^{h}\\ =5 \cdot 5^{h}-3\cdot (-1)^h \\ =5^{h+1} +3 \cdot (-1)^{h+1}$$
Replace $h+1$ by $h+2$ and you'll see that he got exactly the same answer as you, written in a different form.
Not everything which looks different IS different.