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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.

Mardymar
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  • Thank you for your answer's guys. I like your approaches, but I am wondering if it would be possible to start with finding the equation for the differences between the sums and deriving the sums from there. It seems like we should be able to do it both ways. – Mardymar Jul 05 '18 at 13:40
  • Btw, I would upvote you all if I could :) – Mardymar Jul 05 '18 at 13:40
  • For anyone looking for an approach that doesn't involve computing $s_n$ first, see https://math.stackexchange.com/questions/2844896/solution-to-sequence-defined-by-s-1-11-and-s-n1-frac23s-n5-that-a . – Mardymar Jul 08 '18 at 21:48

4 Answers4

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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

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

hamam_Abdallah
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  • I already got the answer for s(n). I did it by finding the difference between each s(n) and summing them. I do like your answer better, though. But is there a way to just find the difference without first finding the sums? – Mardymar Jul 04 '18 at 23:31
  • It is good to know that $10-11=-1$$ – hamam_Abdallah Jul 05 '18 at 00:13
  • If you have time, would you mind giving a little more info on the leap you made on your 'and' statement. I don't see how this is derived. Thanks. – Mardymar Jul 05 '18 at 00:16
  • @Mardymar If $(u_n)$ is geometric, then $u_n=q^{n-1}u_1$ or $u_{n+1}=q^nu_1$. – hamam_Abdallah Jul 05 '18 at 22:59
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HINT

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

Jay Zha
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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.

Ovi
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  • Very clear, thank you. Please see my comment I added to the question. – Mardymar Jul 05 '18 at 13:41
  • @Mardymar I will think about it but right now I don't see any way to find $s_{n+1}-s_n$ before finding $s_n$ first. But I have never studied recursive sequences, I just know one or two tricks, so there may be a very easy way to do it. – Ovi Jul 05 '18 at 17:17
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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,$$S_n=10+\left(\frac{2}{3}\right)^{n-1}.$$

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