Sequences, Series

**Viral** · October 2nd 2009, 01:25 PM

4. Prove that the terms of the sequence $U_n = n^{2}-10n+27$ are positive. For what value of $n$ is $U_n$ smallest?

How do I do this?

**gs.sh11** · October 2nd 2009, 01:36 PM

Well first, n = 0 for $U_n$ to be the lowest, because any negative numbers squared would turn out positive, 0 is smaller than any positive number, so to answer your question n = 0. Im not sure how to prove that they are positive, but I think it has something to do with squaring, anytime you square a negative it becomse positive, and -10 times any negative number will become positive.

**TheEmptySet** · October 2nd 2009, 01:38 PM

Originally Posted by Viral

4. Prove that the terms of the sequence $U_n = n^{2}-10n+27$ are positive. For what value of $n$ is $U_n$ smallest?

How do I do this?

First we need to complet the square

$n^2-10n+27=n^2-10n+25+2=(n-5)^2+2$

Since $(n-5)^2 \ge 0$ we know that

$(n-5)^2+2\ge 0+2=2>0$ and since this is a parabola its mininum occurs at the vertex when n=5 you get 2.

**Viral** · October 2nd 2009, 01:41 PM

Can you expand on:

$ (n-5)^2+2\ge 0+2=2>0 $

a bit please. I understood the completing the square bit fine.

**TheEmptySet** · October 2nd 2009, 01:48 PM

Originally Posted by Viral

Can you expand on:

$ (n-5)^2+2\ge 0+2=2>0 $

a bit please. I understood the completing the square bit fine.

If you square any real number it is always non negative

so $(n-5)^2$ is always bigger than or equal to zero.

i.e the smallest it can ever be is 0 and that occurs when n=5

so the smallest value your squence can take it 2.

**Viral** · October 2nd 2009, 01:51 PM

Thanks I got that now. How would I work out the first part of the question though? (the proof)

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