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It's been a lot time but I have got no response, so asking separately though I know it seems a noob thing to ask but what to do, I'm unable to wrap my head around it :/ .

In this question both the answers find the value of $T_n$ as: $$T_n=4T_{n-1}$$ $$\Rightarrow T_n=4^{n-1}T_1$$ But I'm unable to get how they moved from former to latter that easily?!
I can slowly, step-by-step arrive at the latter using the former but I want to be able to "take that leap" (Inception pun :p).

Please help so I can repeat such a thing for other variants too easily with no doubts in mind.

This is how see it to able to figure it out:
$T_1=k$
$T_2=T_1\cdot r=kr$ $T_2=T_2⋅r=kr^2$
$\vdots$
$T_n=k⋅r^{n-1}$

But I have solved a lot of GPs before and still whenever I look at it as $T_n=r\cdot T_{n-1}$ I'm forced to take a round about way :/ .

InanimateBeing
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    If you understand why the claim holds, then the only issue is really just ... being used to it and exposed to it. At some point you might use that result so much that instead of needing to justify it each time to yourself, you just remember it off the bat – PrincessEev Sep 12 '22 at 17:35
  • @PrincessEev is there really no trick to it? or perhaps some perspective I'm missing? I think I'm definitely lacking some clarity somewhere. Though I haven't been able to figure it out myself yet. – InanimateBeing Sep 12 '22 at 17:38
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    By substitution: $T_n=4T_{n-1}=4^2T_{n-2}=4^3T_{n-3}=\cdots$. After $k$ iterations, you end up with $T_n = 4^kT_{n-k}$. Now let $k=n-1$ to recover $T_n$ in terms of $T_1$. – user170231 Sep 12 '22 at 17:43
  • If at each step you multiply by $4$ then in $n$ steps you multiply by $4$ that many times, so multiply by $4 \times 4 \times 4 \times \cdots \times 4 = 4^n$. – Ethan Bolker Sep 12 '22 at 17:48
  • @user170231 thank you, I too thought of this, but I rejected it because I was confused as to why are putting k=n-1 instead of k=n-2? Also, I wasn't finding it an "appealing"-enough approach (something that strikes you as "That's it!"-kind of thing) – InanimateBeing Sep 12 '22 at 17:51
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    If we take $k=n-2$, we get$$T_n = 4^{n-2} T_{n-(n-2)} = 4^{n-2} T_2$$which we could use just as effectively, with the extra brief step of computing $T_2=4T_1$. – user170231 Sep 12 '22 at 17:53
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    For me, I do the same substitution as above ($T_n = 4T_{n-1} = \ldots = 4^kT_{n-k}$), then spend equal or more time to double check and eliminate any off-by-one error (so I don't instead write $T_n = 4^nT_1$). Sometimes things that appears easy to write may have more thoughts hidden. – peterwhy Sep 12 '22 at 17:58
  • We put $k=n-1$ into $T_n = 4^kT_{n-k}$, because then the RHS is in terms of $T_1$, which is given. If instead the question gives $T_0$, then we instead substitute $k=n$ to get $T_n = 4^nT_0$. – peterwhy Sep 12 '22 at 18:02

4 Answers4

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Use induction.
$T_2=4T_1$.

Assume $T_n=4^{n-1}T_1$.
$T_{n+1}=4T_n=4\cdot 4^{n-1}T_1=4^n T_1$.

InanimateBeing
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ryaron
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  • It's better to use \cdot which is $\cdot$ instead of $$. Sometimes we go for \times, that is $\times$ instead. (Never $$ though). – InanimateBeing Sep 12 '22 at 18:05
  • Thank you for the answer. +1. Induction might be helpful but I wanted more quicker way of achieving the result while induction requires a little more time. – InanimateBeing Sep 13 '22 at 17:40
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By substitution:

$$\begin{align*} T_n &= 4 T_{n-1} \\ &= 4(4T_{n-2}) = 4^2 T_{n-2} \\ &= 4(4^2 T_{n-3}) = 4^3 T_{n-3} \\ &~\vdots \\ &= 4^k T_{n-k} \end{align*}$$

Let $k=n-1$ to recover $T_n$ in terms of $T_1$. It follows that

$$T_n = 4^{n-1} T_{n-(n-1)} = 4^{n-1} T_1 = 4^n$$

user170231
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Slightly differently: Multiplicative Telescoping: $$T_2=4T_1$$ $$T_3=4T_2$$ $$T_4=4T_3$$ $$T_5=4T_4$$ $$........$$ $$........$$ $$T_{n-1}=4T_{n-2}$$ $$T_n=4T_{n-1}$$ Multiply all these equations observe cancellations on left and right to get $$T_n=4^{n-1}T_1.$$

Z Ahmed
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I like to think about this particular type of recurrence in terms of how exponents are defined. Concretely, we can define $$a^0 = 1,\quad a^{n} = a^{n-1} \cdot a,\ n\in\mathbb N.\tag{1}$$ Note that this is a recurrence relation and when we write $a^n$ we do mean a member of the unique sequence that solves $(1)$.

Now, let's consider recurrence $$T_{n+1} = a\, T_{n},\ n\in\mathbb N.\tag{2}$$ Assuming $T_1\neq 0$, define sequence $a_n = T_{n+1}/T_1$. Dividing $(2)$ by $T_1$ we get $$a_0 = 1,\quad a_n = a_{n-1} \cdot a,\ n\in\mathbb N$$ so $a_n$ solves $(1)$ and by uniqueness it follows that $a_n = a^n$. Multiplying by $T_1$ now gives $$T_{n+1} = a^n\, T_1,\ n\in\mathbb N.$$

Of course, if $T_1 = 0$, then $T_n = 0,\forall n\in\mathbb N,$ and the formula still holds.

If you understand that $(1)$ and $(2)$ are almost the same and that they differ only by multiplication by a constant, whenever you see a recurrence of the type $(2)$, you can immediately conclude that $T_n = C\,a^n$ for some constant $C$. Plugging in $n = 1$ will give you that $C = T_1/a$, so $T_n = a^{n-1}\, T_1.$

Ennar
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  • Why did you define $a_n=T_{n+1}/T_1$? – InanimateBeing Sep 13 '22 at 17:56
  • Because the only difference between $(1)$ and $(2)$ is that $T_1$ might not be $1$. I.e., the only difference is in initial condition (and the sequence is shifted by 1). So, when we divide by $T_1$, we get a sequence that solves $(1)$. @InanimateBeing – Ennar Sep 13 '22 at 17:59
  • @InanimateBeing I have edited my answer adding the actual answer to your question "how to take that leap". The long introduction is why that leap works. – Ennar Sep 13 '22 at 18:31
  • @InanimateBeing, don't worry a bit about it. My answer is maybe not as easily accessible as others, but I've attempted my best to not necessarily give you a computation trick, but understanding of what's going on. – Ennar Sep 14 '22 at 01:20
  • Thank you for the answer. +1. Though I think this is a bit of a round-about way of going about it. But it's fine too. – InanimateBeing Sep 14 '22 at 04:08
  • @InanimateBeing saying this is round-about is like saying that coding in assembler is round-about compared to coding in C. – Ennar Sep 14 '22 at 11:30