Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$?

Question

How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?

Answer 1

Hint: $$a_{n+1} - 3a_n = 5 \cdot 3^{n+1} = 3 (a_n-3a_{n-1})$$

Can you start from here?

$$a_{n+1} - 6a_n + 9a_{n-1} = 0 \implies \frac{a_{n+1}}{3^{n+1}} - 2\frac{a_n}{3^n} + \frac{a_{n-1}}{3^{n-1}}=0$$ Let $a_n = 3^n b_n$ then $$b_{n+1} - 2b_n + b_{n-1} = 0 \implies b_{n+1} - b_{n} = b_n-b_{n-1} = \cdots = b_1 - b_0 = 7 - 2 =5$$ Therefore $b_n=b_0+5n=5n+2, a_n = 3^n (5n+2). \blacksquare$

Answer 2

Put $$a_n=5.3^n.b_n$$

then

$$5.3^n.b_n=3.5.3^{n-1}.b_{n-1}+5.3^n$$

thus

$$b_n=b_{n-1}+1$$

So, $$b_n=n+2,\; \; and \; \; a_n=5.3^n(n+2)$$

Answer 3

Let $ n\in\mathbb{N} $, we have for any $ k\leq n-1 $ the following : $$ a_{n-k}=3a_{n-k-1}+5\times 3^{n-k} $$

Thus : \begin{aligned}3^{k}a_{n-k}&=3^{k+1}a_{n-k-1}+5\times 3^{n}\\ \Longrightarrow \sum_{k=0}^{n-1}{\left(3^{k}a_{n-k}-3^{k+1}a_{n-k-1}\right)}&=5n\times 3^{n}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{n}-3^{n}a_{0}&=5n\times 3^{n}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{n}&=3^{n}\left(5n+2\right)\end{aligned}

Since $ n $ is arbitrary, the result stands for all $ n\in\mathbb{N} $.

Hope that helps.

Answer 4

When you have a first-order recurrence like this, you can often simply ‘unwind’ it:

$$\begin{align*} a_n&=3a_{n-1}+5\cdot3^n\\ &=3(3a_{n-2}+5\cdot3^{n-1})+5\cdot3^n\\ &=3^2a_{n-2}+2\cdot5\cdot3^n)\\ &=3^2(3a_{n-3}+5\cdot3^{n-2})+2\cdot5\cdot3^n\\ &=\color{red}{3^3a_{n-3}+3\cdot5\cdot3^n}\\ &\;\;\vdots\\ &=\color{blue}{3^ka_{n-k}+k\cdot5\cdot3^n}\\ &\;\;\vdots\\ &=3^na_0+n\cdot5\cdot3^n\\ &=3^n(2+5n) \end{align*}$$

By the time you get to the red line, the pattern should be pretty clear; I’ve expressed it in general terms in the blue line. Then you can jump to the end and simplify to get a closed form for $a_n$. The step from the red to the blue line isn’t rigorous, so properly speaking you should prove by induction on $n$ that the formula $a_n=3^n(5n+2)$ really does satisfy the recurrence with the right initial condition, though in many cases, like this one, the pattern really is very clear.

Answer 5

general sequence is $a_n= (3^n)a_0 +n \times 5\times 3^n$ substitute $a(n-1)$ for $a(n-2)$ and check the pattern you can easily figure it out.