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 22nd May 2013, 08:21 PM #1 (permalink) New SGClubber   Posts: 20 Join Date: May 2013 Likes: 0 Thanked 1 Time in 1 Post Gender: Recurrence relation given. Find formula for nth term. What is the best way? Tweet A sequence u0, u1, u2, ... is such that u(n+1) = 0.9u(n) + 90. Show that u(n) = (0.9^n)(u0 - 900) + 900. This question is quite easy. I'm only wondering whether I have the most efficient technique. My solution is shown below. Would appreciate if someone could show me a more efficient way of doing this particular question, assuming there is one. (Not counting math induction.) What I do is find the pattern: u0 = u0 u1 = (0.9)u0 + 90 u2 = (0.9^2)u0 + (0.9)90 + 90 u3 = (0.9^3)u0 + (0.9^2)90 + (0.9)90 + 90 . . . u(n) = (0.9^n)u0 + (0.9^n-1)90 + (0.9^n-2)90 + ... + (0.9)(90) + 90 After this it's easy, because the first term is (0.9^n)u0 and the rest are just n terms of a G.P. So it's easy to sum everything and the correct answer is obtained. But is there a "cleaner" (more elegant) way of finding u(n) for this particular example? Appreciate any tips. Sponsors:
 31st May 2013, 04:49 PM #2 (permalink) Cool SGClubber   Posts: 129 Join Date: Nov 2011 Likes: 7 Liked 11 Times in 10 Posts Gender: Re: Recurrence relation given. Find formula for nth term. What is the best way? Tweet Actually, i think your method is the most straightforward way for most people/students. Mine is not... For me i considered: $u_{n}-u_{n-1}=0.9(u_{n-1}-u_{n-2})$ Let: $a_{n}=u_{n}-u_{n-1}$ So, i will get: $a_{n}=0.9a_{n-1} \Rightarrow \frac{a_n}{a_{n-1}}=0.9$ So i will get a GP in general with: $a_{n}=0.9^{n-1}(u_1-u_0)=0.9^{n-1}(90-0.1u_0)$ Now consider: $\sum_{i=1}^{n}a_i=\sum_{i=1}^{n}(u_i-u_{i-1})$ I will than use gemoteric series and method of difference to produce: $\frac{(90-0.1u_0)(1-0.9^n)}{0.1}=u_n-u_0$ Upon simplification: $u_n=900+0.9^n(u_0-900)$ You should kinda realise by now that this method is very similar to the one proposed. Last edited by megaman123; 31st May 2013 at 04:51 PM.
 Members who Liked this post by megaman123: workingtoohard (2nd June 2013)
 5th June 2013, 09:48 PM #3 (permalink) New SGClubber   Posts: 20 Join Date: May 2013 Likes: 0 Thanked 1 Time in 1 Post Gender: Re: Recurrence relation given. Find formula for nth term. What is the best way? Tweet Eh, thanks for this! Took me a while to see how you got that first line so quickly: $u_{n}-u_{n-1}=0.9(u_{n-1}-u_{n-2})$ but once I saw it, it was quite obvious. I've seen the MoD being applied in this way before, so I could follow the rest very quickly. It's just what I was looking for, more or less! Very elegant, this way. You one cool dude la. But I doubt I can teach this to my students cos it would be too much for the average student. But anyway, from what I can see, you can use this method to find u(n) whenever the recurrence relation is of the linear form form u(n) = Au(n-1) + B. That's great, I'll use this method the next time. Thanks again!

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