I have the following recurrence relation:
$T(n) = \dfrac{2}{5} \times T(n-1) + \dfrac{3}{5} \times T(n-2)$
with base cases:
$T(0) = 0$ & $T(1) = 1$
I need help proving, by induction, that this relation is given by:
$T(n) = - \dfrac{(-3)^n - 5^n}{8 \times 5^{n-1}}$
PS: Apologies for the poor formatting. I'm brand new here.
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When $n=0$, we have $$T(0) = 0 = -\frac{(-3)^0-5^0}{8\cdot 5^{0-1}},$$ which agrees with the formula, starting the induction.
Now for the inductive step, suppose $T$ is given by your formula for all $0\leqslant k\leqslant n$. We need to show that $T(n+1)$ is also given by your formula. Indeed, \begin{align*} T(n+1) &= \tfrac 25T(n)+\tfrac35T(n-1)\\[3pt] &= \frac25\Big(\frac{(-3)^n-5^n}{8\cdot5^{n-1}}\Big) + \frac35\Big(\frac{(-3)^{n-1}-5^{n-1}}{8\cdot5^{n-2}}\Big)\quad(\text{induction hypothesis})\\[3pt] &=\frac{2(-3)^n-2(5^n)+5\cdot3(-3)^{n-1}-5\cdot3\cdot5^{n-1}}{8\cdot5^n}\\[3pt] &=\frac{2(-3)^n-2(5^n)-5\cdot(-3)(-3)^{n-1}-3\cdot5^n}{8\cdot5^n}\\[3pt] &=\frac{-3(-3)^n-5(5^n)}{8\cdot5^n}=\frac{(-3)^{n+1}-5^{n+1}}{8\cdot5^{(n+1)-1}}, \end{align*} which shows that your formula holds for the $n+1$ case.