Finding $T_{200}$ of given series In a certain series, the $n$th term, $T_n=4T_{n-1} + n – 1$. If $T_1$ = 4, then find the value of $T_{200}$.
I tried the following:
$T_n-T_{n-1}=3T_{n-1} + n – 1$
$T_{n-1}-T_{n-2}=3T_{n-2} + n – 2$
$T_{n-2}-T_{n-3}=3T_{n-3} + n – 3$
Then I proceeded to calculate the difference of the first $2$ and the last $2$ to get:
$$T_n-3T_{n-1}+3T_{n-2}-T_{n-3}=3(T_{n-1} -T_{n-3})$$
Well $\dots$ basically I didn't reach anywhere (neither LHS nor RHS is in friendly form). Then I also tried taking $n^{th}$ order differences and then I got a GP at $3^{rd}$ order difference but still didn't know what to do from there (I'm only familiar with APs of $n^{th}$ order difference).
How to solve this problem? Please help.
Answer (and its format):

 $$\frac {10(4^{200})-601}9$$

Edit:
Kindly suggest such that even if the question is tweaked a bit, I may be able to solve that too. For example:
$$T_n=3T_{n-1} + n – 1\text { and } T_1 = 3 \text { then } T_{100}=?$$
 A: By the question, we know $T_n+\frac{n}{3}+\frac{1}{9}=4(T_{n-1}+\frac{n-1}{3}+\frac{1}{9})$, so
$T_{200}+\frac{200}{3}+\frac{1}{9}=4^{199}(t_1+0+\frac{1}{9})=4^{199}(4+\frac{1}{9})$, thus, $T_{200}=4^{200}+\frac{4^{199}}{9}-\frac{601}{9}$.
A: The idea is to transform the original sequence $\{T_n\}$ to a geometric sequence by absorbing the term $n - 1$ into the transformed sequence. To this end, we are trying to rewrite the original recurrence relation as
\begin{equation*}
T_n + an + b = 4(T_{n - 1} + a(n - 1) + b),
\end{equation*}
where $a$ and $b$ are to be determined by
\begin{align*}
\begin{cases}
4an - an = n \\[1em]
4b - 4a - b = -1
\end{cases}.
\end{align*}
Therefore, $a = 1/3, b = (4a - 1)/3 = 1/9$.  Hence
\begin{align*}
T_n + \frac{1}{3}n + \frac{1}{9} = \left(T_1 + \frac{1}{3} + \frac{1}{9}\right)4^{n - 1}.  
\end{align*}
$T_{200}$ then follows easily.
A: Given
$$T_n = 4T_{n-1} + n - 1$$
by substitution we have
$$\begin{align*}
T_n &= 4 T_{n-1} + n - 1 \\[1ex]
&= 4^2 T_{n-2} + (n-1) + 4 (n-2) \\[1ex]
&= 4^3 T_{n-3} + (n-1) + 4 (n-2) + 4^2 (n-3) \\[1ex]
&~\vdots \\[1ex]
&= 4^k T_{n-k} + (n-1) + 4 (n-2) + 4^2 (n-3) + \cdots + 4^{k-1} (n-k) \\[1ex]
&= 4^k T_{n-k} + \sum_{i=1}^k 4^{i-1} (n-i)
\end{align*}$$
When $k=n-1$, we have
$$\begin{align*}
T_n &= 4^{n-1}T_{n-(n-1)} + \sum_{i=1}^{n-1} 4^{i-1} (n-i) \\[1ex]
&= 4^{n-1}T_1 + n \sum_{i=1}^{n-1} 4^{i-1} - \sum_{i=1}^{n-1} i \cdot 4^{i-1} \\[1ex]
&= 4^n + \frac{n(4^n-4)}{12} - \frac{4^{n+1} - 3n\cdot4^n - 4}{36} \\[1ex]
&= -\frac{4^{n+1}}{36} + 4^n - \frac n3 + 9
\end{align*}$$
and from here you can evaluate $T_{200}$. See here for methods on computing the geometric sums.
