trying to help son via email with problem!

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drb2748

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Oct 16, 2009
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It's been almost 20 years since I even looked at an algebra problem but I told my son, who is flunking math, that I would help him. I am stuck. Know it is something simple but just not getting it. This is what I have so far:

Problem: U1=40 Un=u(n-1)+13 where n is equal or greater than 1.

Un = u(n - 1) + 13

If U1 = 40 then U = 40 (U x 1 = 40 x 1)

40n = un - u1 + 13

tried dividing by n and making n=1 and no go.

what am I missing?
 
What are you needing to do? (I see two expressions, but no instructions for what you're supposed to do with them.)

Thank you! :wink:
 
to solve the equation of: Un=u(n-1)+13

where n is equal or greater than 1
U1=40
 
This does not appear to be something you can divide by n. It looks to be a recurrence relation. If I am looking at it correctly.

Are those n's subscripts?.

\(\displaystyle U_{n}=U_{n-1}+13, \;\ U_{1}=40\)

Or perhaps it is \(\displaystyle U(n)=U(n-1)+13\). Means the same thing.

If this is the case, you can not just divide by n. It is not that kind of problem.

If we know \(\displaystyle U_{1}=40\), then if we sub in n=1, we get \(\displaystyle U_{1}=U_{0}+13\) and get \(\displaystyle 40=U_{0}+13\Rightarrow U_{0}=27\)

Is that all he supposed to do?. The general form for this would be 27+13n. It is just an arithmetic series with common difference of 13. Plug in n=whatever and get the value for that n.

n=0=27+13(0)=27

n=1, 27+13(1)=40

n=2, 27+13(2)=53

and so on.

See?. We're just adding 13 each time.
 
I had gone that route too and checked with my son to make sure exactly how the equation was written and you have 2 different un's ; you have a Un and a un. Doesn't that make a difference?
 
I have a Un and a un?. I don't understand. Anyway, no, it does not make a difference.
 
so Un=un? they why distinguish between the 2 by capitalizing one of them?
 
Hello, drb2748!

What do you mean by "solve the equation"?

\(\displaystyle \text{Problem: }\; U_1 =40,\;\;U_n \:=\:U_{n-1}+13,\;\text{ where }n \geq 1\)
Click to expand...

It's kind of obvious, isn't it?

Each term is 13 more than the preceding term.
The sequence starts at 40 and "goes up by 13".

. . \(\displaystyle \begin{array}{c|c} n & U_n \\ \hline 1 & 40 \\ 2 & 53 \\ 3 & 66 \\ 4 & 79 \\ \vdots & \vdots \end{array}\)

\(\displaystyle \text{The general term is: }\;U_n \;=\;27 + 13n\)

 
Thank you all for your help. Now that I know what he is doing I can help him.
 
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