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A sequence is defined by the recurrence relation $u_{n+1} = 0.2u_n + 9$, ${u_5 = 11}$. What is the value of ${u_3}$?

enter image description here

I have not encountered a problem like this when only one value for n is provided. Normally I am given a couple of values for n and I can solve for u by solving simultaneous equations.

I thought I could put 11 into the equation and substitute 5 for n and find u.

${11 = 0.2u(5) + 9}$

=> ${u = 2}$

${u_3 = 0.2(3)(2) + 9} = 10.2$

But the answer is incorrect.

dagda1
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  • what is $u_{n+1}$? – Dr. Sonnhard Graubner Dec 26 '15 at 20:05
  • Why is the $u$ subscripted in ${u_{n+1} = 0.2_u + 9}$. Should it be $u_n$? – Mike Pierce Dec 26 '15 at 20:07
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    the question is exactly how it is asked in the Scottish higher exam of 2015 so give me one good reason why this has been put on hold – dagda1 Dec 27 '15 at 16:38
  • The notation doesn't make sense. What does your subscript $u$ mean? You write the line ${11 = 0.2u(5) + 9}$ saying that you plugged in $5$ for $n$, but there was no $n$ there in the original expression. Your "recurrence relation" isn't even a recurrence relation (what is that mysterious $u$ supposed to be?). Also, you decided to omit the fact that this question came from the "Scottish higher exam", which could be important context to include in your post. You also could've provided a link to the original question, which would remove any ambiguity for readers. – Mike Pierce Dec 27 '15 at 17:47
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    Notice that you copied the problem here incorrectly. You typed ${u_{n+1} = 0.2_u + 9}$ while the actual question in the image has $u_{n+1} = 0\cdot 2u_n + 9$. – Mike Pierce Dec 27 '15 at 20:19

1 Answers1

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Note that I have edited this answer since OP posted an image of the original question.

According to the recurrence relation, we have $u_5 = 0.2u_4 + 9$ and $u_4 = 0.2u_3 + 9$. Then we have $$\begin{align} u_5 &= 0.2u_4 + 9 \\ u_5 &= 0.2(0.2u_3 + 9) + 9 \\ (11) &= 0.2(0.2u_3 + 9) + 9 \\ 2 &= 0.2(0.2u_3 + 9) \\ 2/(0.2) &= (0.2u_3 + 9) \\ 10 &= 0.2u_3 + 9 \\ 1 &= 0.2u_3 \\ 1/(0.2) &= u_3 \\\\ 5 &= u_3 \end{align}$$

Mike Pierce
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