Finding formulas for the terms in a recursion defined by $s_1 = 11$ and $s_{n+1}=\frac23(s_n+5)$

Question

Say you have a recursion defined by: $$\begin{align}s_1 &= 11 \\[4pt] s_{n+1} &= \frac23 ( s_n + 5 ) \end{align}$$ I am trying to find an equation that allows the user to put in $n$ and get back the difference change: $s_{n+1} - s_{n}$. I already know this formula is:

$$\frac13\left( \frac23 \right)^{ n - 1 }( 10 - s_1 )$$ which I found by plugging in numbers and looking at the results. I don't understand how to derive this equation, though. I tried a few approaches, but none of them led me anywhere.

Answer 1

hint

Let us look for $a$ such that

$$s(n+1)-a=\frac 23(s(n)-a)$$

then

$$a-\frac 23a=\frac{10}{3}$$ or $$a=10$$

thus $$s(n+1)-a=\frac 23(s(n)-a)$$ and $$s(n+1)=(\frac 23)^n(s(1)-a)+a=(\frac 23)^n+10$$

finally

$$s(n+1)-s(n)=\frac{-1}{3}(\frac 23)^{n-1}$$

Answer 2

HINT

$$(S_{n+1} - 10 )= \frac{2}{3} (S_n - 10)$$

Answer 3

Expanding on the other two answers:

We are at $(s_{n+1} - 10 )= \dfrac{2}{3} (s_n - 10)$. The point of bringing it to this form is that now we can define a new sequence, $u_n = s_n-10$. Now the equation above turns to $u_{n+1} = \dfrac 23 u_n$, with $u_1 = s_1-10 = 1$.

So $u_1 = 1$, $u_2 = \dfrac 23, u_3 = \left(\dfrac 23\right)^2, u_4 = \left(\dfrac 23\right)^3$, and so on. So it appears that $u_n = \left(\dfrac 23\right)^{n-1}$. Now to prove that our solution is correct, we just plug in $u_n$ into $u_{n+1} = \dfrac 23 u_n$ and make sure that both sides match:

RHS = $\left( \dfrac 23 \right)^n$

LHS = $ \dfrac 23 \cdot \left( \dfrac 23 \right)^{n-1} = \left( \dfrac 23 \right)^n$

This proves that our solution is indeed correct.

Now $u_n = s_n-10$, so $\left( \dfrac 23 \right)^{n-1}= s_n-10$, thus $s_n = \left( \dfrac 23 \right)^{n-1}+10$.

And now,

$$s_{n+1}-s_n = \left(\left( \dfrac 23 \right)^n+10\right) - \left(\left( \dfrac 23 \right)^{n-1}+10 \right) = \left( \dfrac 23 \right)^n - \left( \dfrac 23 \right)^{n-1} = -\dfrac 13 \cdot \left( \dfrac 23 \right)^{n-1}$$

Let me know if there is anything that needs clarification.

Answer 4

Solution

By the method of undetermined coefficients，we may assume that $$pS_{n+1}+q=k(pS_n+q).$$ Thus, $$pS_{n+1}-kpS_n+q-kq=0.$$

In comparison with the recursion formula $3s_{n+1}-2s_n-10=0$,we have $$p=3,kp=2,q-kq=-10.$$ Hence, $$p=3,q=-30,k=\frac{2}{3}.$$ This shows that $$3S_{n+1}-30=\frac{2}{3}(3S_n-30),$$namely,$$S_{n+1}-10=\frac{2}{3}(S_n-10).$$ This shows that $\{S_n-10\}$ is a geometric sequence. It follows that $$S_{n}-10=(S_1-10)\left(\frac{2}{3}\right)^{n-1}=\left(\frac{2}{3}\right)^{n-1}.$$As a result,$$S_n=10+\left(\frac{2}{3}\right)^{n-1}.$$