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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: $$T_n=3T_{n-1} + n – 1\text { and } T_1 = 3 \text { then } T_{100}=?$$

InanimateBeing
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3 Answers3

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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, $T_{200}=4^{200}+\frac{4^{199}}{9}-\frac{601}{9}$.

qiujie qiao
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  • Thank for the answer. I have edited my post. Could you please suggest how I can formulate such a grouping (quickly) if the question is tweaked a bit (as now the edited question contains)? – InanimateBeing Jul 22 '22 at 15:36
  • You can assume $T_n+an+b=3(T_{n-1}+a(n-1)+b)$, and solve the coefficients $a$ and $b$ by the question, then, you could solve questions like this. The most important thing is to give a formula like $c_n=kc_{n-1}$, you just need to transform the formula in question into the simple formula $c_n=kc_{n-1}$ and find what $c_n$ is. – qiujie qiao Jul 23 '22 at 09:45
  • Thank you. Also, could you please tell that how did we know that we need to transform the original sequence to a geometric sequence (and that other terms will fall in place accordingly)? – InanimateBeing Jul 23 '22 at 13:19
  • I think it's depend the type of $T_n$ and $T_{n-1}$ that the question gives, for example, if the type is $T_n$ and $k T_{n-1}$, then we could guess the formula could be transformed to $c_n=k c_{n-1}$; if the type is $T_n$ and $T_{n-1}$, then we could guess the formula could be transformed to $c_n= c_{n-1}+m$, if the type is $T_n$ and $T_{n-1}^2$, then we could guess the formula could be transformed to $c_n= c_{n-1}^2$. What you guess is all depend the type of the $T_n$ and $T_{n-1}$, that's my opinion, I hope my thoughts could help you, wish you best. – qiujie qiao Jul 24 '22 at 00:37
  • They sure did help me a lot. Thank you. Already upvoted. – InanimateBeing Jul 24 '22 at 04:23
  • Hey, sorry for being so late but I realised that I accepted the answer wrongly. Could you please edit your answer to include whatever you have answered in comments to make it better? – InanimateBeing Aug 29 '22 at 13:08
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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.

Zhanxiong
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  • Thank you for the answer. From the first 2 sentences it seems you have also established that $a_n=a_{n-1}$. Is there a typing error or am I not getting something? – InanimateBeing Jul 22 '22 at 15:30
  • @InanimateBeing What has been established is $a_n = 4a_{n - 1} - 1$ -- what makes you think that $a_n = a_{n - 1}$ was established??? By the way, your edited new question is trivial if you really understand the above process. – Zhanxiong Jul 22 '22 at 15:45
  • I know edit is trivial for you. I said above statement because $T_n + kn = 4(T_n + kn) - 1$ Here you have taken $(T_n + kn)$ on the LHS as $a_n$ and $(T_n + kn)$ on the RHS as $a_{n-1}$. Please clear where I'm wrong. – InanimateBeing Jul 22 '22 at 15:54
  • I, though apparent, hesitated to call it a typo as once you correct that then other doubts arise. Now $a_{n-1}$ wouldn't be equal to $T_{n-1}+n/3$. Could you please proofread the whole and check if it matches the answer. If have no problem on continuing on my own and did it first time too. – InanimateBeing Jul 22 '22 at 16:05
  • @InanimateBeing My apology. Edited the answer. – Zhanxiong Jul 22 '22 at 16:09
  • How did we know that we need to transform the original sequence to a geometric sequence (and that other terms will be absorbed)? – InanimateBeing Jul 24 '22 at 18:44
  • @InanimateBeing Let $c_n = T_n + n/3 + 1/9$, then the answer gives you $c_n = 4c_{n - 1}$, which is a geometric sequence, which implies its $n$-th term is $c_n = c_1 4^{n - 1}$. – Zhanxiong Jul 25 '22 at 04:22
  • Finally, in your final equation how were you directly able to write/bring T1, I'm unable to get the logic? (One can look at it as going from $c_2$ to $c_{200}$ but this thought process consumes a lot of time, so how were you able to skip this train of thought?) – InanimateBeing Jul 25 '22 at 06:58
  • As I already explained, can't you see $c_1 = T_1 + 1/3 + 1/9$, by the notation I used in last comment? Honestly it is a fair straightforward substitution, why it looks so burdernsome to you? – Zhanxiong Jul 25 '22 at 11:52
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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.

user170231
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  • Thank you. Wow! Didn't expect too see it solved this way as I tried my best to form such a series but wasn't successful. Glad to know that this option is available too. +1. Thank you. – InanimateBeing Sep 12 '22 at 18:14