I have encountered an induction problem which I don't understand. What I don't understand is what it is asking me to prove. I don't want a solution. The problem is:
If $u_1=5$ and $u_{n+1}=2u_n-3(-1)^n$, then $u_n=3(2^n)+(-1)^n$ for all positive integers.
Am I supposed to prove $u_{n+1}=2u_n-3(-1)^n$ or $u_n=3(2^n)+(-1)^n$ is true for all positive integers?
On
$u_{n+1}=2u_n-3(-1)^n$ is your recurrence, specifying on how to obtain the $n+1$-st term $u_{n+1}$ from the $n$-th term $u_n$. Together with a start value, $u_1=5$, this stepwise determines the sequence completely.
What the problem is trying to establish, is to show that in general, for any $n$, you can obtain the $n$-th term directly(without evaluating all $u_k$'s with $k<n$ before) via the formula $u_n=3(2^n)+(-1)^n$. That this formula holds can be shown using induction.
On
You are given $u_{n+1}=2u_n-3(-1)^n$. This would allow you to compute the entire series. For example $$u_2=2u_1-3(-1)^1=2\cdot 5+3=13\\ u_3=2\cdot 13-3(-1)^2=23$$ and so on. You are supposed to prove the last statement. It claims $$u_1=3(2^1)+(-1)^1=5\\u_2=3(2^2)+(-1)^2=13\\u_3=3(2^3)+(-1)^3=23\\\vdots$$
You are supposed to prove $u_n=3(2^n)+(-1)^n$.
$u_1=5$ and $u_{n+1}=2u_n-3(-1)^n$ are the conditions you are supposed to make use of.