Find $a_n$ via $a_n - a_{n-1}=3^n$

Question

I remember the middle school taught how to find the closed-form expression of $a_n$ given $$ a_0=1 $$ $$ a_n - a_{n-1}=3^n $$ What's name of such sequence? What's general approach to obtain $a_n$?

ps. the sequence is $1,4,13,40, \dots$, i.e. $$ a_n=\sum_{i=0}^n 3^i $$ but in this case I don't regard $\sum$ is a closed-form expresstion.

Answer 1

I generalize a tad, setting

$a_0 = a, \tag 1$

$a_n - a_{n - 1} = b^n; \tag 2$

then

$a_1 - a_0 = a_1 - a = b, \tag 3$

$a_1 = a + b; \tag 4$

$a_2 - a_1 = a_2 - (a + b) = b^2, \tag 5$

$a_2 = a + b + b^2; \tag 6$

I make the inductive hypothesis:

$a_k = a + \displaystyle \sum_1^k b^j; \tag 7$

then

$a_{k + 1} = a_k + b^{k + 1} = a + \displaystyle \sum_1^k b^j + b^{k + 1} = a + \sum_1^{k + 1}b^j; \tag 7$

it follows by induction that for general $n \in \Bbb N$,

$a_n = a + \displaystyle \sum_1^n b^j = a + \dfrac{b^{n + 1} - b}{b - 1}; \tag 8$

taking

$a_0 = a = 1, \; b = 3, \tag 9$

we find

$a_n = 1 + \dfrac{3^{n + 1} - 3}{2} = \dfrac{3^{n + 1} - 1}{2}. \tag{10}$

I'm not sure what the name of this sequence is, but it's called a "geometric something-or-other", I'll warrant.

Answer 2

Hint

$$a_n = 3^n + a_{n-1} \\ a_{n-1} = 3^{n-1} + a_{n-2} \\ a_{n-2} = 3^{n-2} + a_{n-3} \\ ...\\ a_1=3^1+a_0$$

Add everything together.

Answer 3

Telescoping

• $a_n = a_0 + \sum_{k=1}^n (a_k - a_{k-1})$ together with geometric sum
• $\sum_{k=0}^nq^k = \frac{q^{n+1}-1}{q-1}$ give immediately

$$a_n = 1 + \sum_{k=1}^n 3^k = \sum_{\color{blue}{k=0}}^n 3^k = \frac{3^{n+1}-1}{3-1} = \frac{3^{n+1}-1}{2}$$