$T_n=4T_{n-1}\Rightarrow T_n=4^{n-1}T_1$ but how?

Question

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 :/ .

Answer 1

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

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

Answer 2

Use induction.
$T_2=4T_1$.

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

Answer 3

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 $$T_n=4^{n-1}T_1.$$

Answer 4

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 $T_n = a^{n-1}\, T_1.$