[Algebra]
Problem: Let $T=7$. Given the sequence $u_{n}$ such that $u_{3}=5, u_{6}=89$, and $u_{n+2}=3 u_{n+1}-u_{n}$ for integers $n \geq 1$, compute $u_{T}$.

Solution: By the recursive definition, notice that $u_{6}=89=3 u_{5}-u_{4}$ and $u_{5}=3 u_{4}-u_{3}=3 u_{4}-5$. This is a linear system of equations. Write $3 u_{5}-u_{4}=89$ and $-3 u_{5}+9 u_{4}=15$ and add to obtain $u_{4}=13$. Now apply the recursive definition to obtain $u_{5}=34$ and $u_{7}=\mathbf{2 3 3}$.
Notice that the given values are both Fibonacci numbers, and that in the Fibonacci sequence, $f_{1}=f_{2}=1, f_{5}=5$, and $f_{11}=89$. That is, 5 and 89 are six terms apart in the Fibonacci sequence, and only three terms apart in the given sequence. This relationship is not a coincidence: alternating terms in the Fibonacci sequence satisfy the given recurrence relation for the sequence $\left\{u_{n}\right\}$, that is, $f_{n+4}=3 f_{n+2}-f_{n}$. Proof: if $f_{n}=a$ and $f_{n+1}=b$, then $f_{n+2}=a+b, f_{n+3}=a+2 b$, and $f_{n+4}=2 a+3 b=3(a+b)-b=3 f_{n+2}-f_{n}$. To compute the final result, continue out the Fibonacci sequence to obtain $f_{12}=144$ and $u_{7}=f_{13}=233$.

Final answer: 233