Find the least value of n in an arithmetic sequence

**Viral** · October 17th 2009, 02:33 PM

Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$

How do I do this?

**Matt Westwood** · October 17th 2009, 02:40 PM

Originally Posted by Viral

Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$

How do I do this?

Furst thing you do is simplify it down by extracting the constants outside:

$\sum_{r=1}^n (4r-3) = 4 \sum_{r=1}^n r - \sum_{r=1}^n 3 = 4 \sum_{r=1}^n r - 3n$ .

Then you know (or ought to know, or be able to find out, or work out) what $\sum_{r=1}^n r$ is.

Then you have an inequality in n to solve. Good luck.

**Viral** · October 17th 2009, 02:42 PM

Hmm, I'm not really sure what you did there =\ . We've never covered that.

**mr fantastic** · October 17th 2009, 02:44 PM

Originally Posted by Viral

Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$

How do I do this?

$\sum_{r=1}^n (4r-3) = 4 \left( \sum_{r=1}^n r\right) - 3n$ and you should know the formula to substitute for $\sum_{r=1}^n r$ . Then solve for smallest positive integer value of n such that it's greater than 2000.

If you need more help please show all your working and say where you get stuck.

**MacstersUndead** · October 17th 2009, 02:44 PM

Originally Posted by Viral

Hmm, I'm not really sure what you did there =\ . We've never covered that.

Sigma sums preserve multiplication by a constant and preserve addition. You can prove this using the definition of sigma sums.

**Viral** · October 17th 2009, 02:46 PM

Is $\sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))$ ?

**MacstersUndead** · October 17th 2009, 02:49 PM

Originally Posted by Viral

Is $\sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))$ ?

No, but $\sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-(3/4))$

**Viral** · October 17th 2009, 02:51 PM

Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.

**Matt Westwood** · October 17th 2009, 02:52 PM

Originally Posted by Viral

Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.

Fair comment. Can you document on this thread everything you've been taught?

(joke)

**Viral** · October 17th 2009, 02:56 PM

To be honest, I would but I don't know how I'd explain it. We've learned how to calculate all terms in a sequence, how to find a, d, n and l. And this is our first look at the summation notation. If there are other methods, and we've been taught it, it should hit me when I see it.

**mr fantastic** · October 17th 2009, 02:56 PM

Originally Posted by Viral

Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.

The first term in the sum is 4 - 3 = 1. Therefore a = 1.

The common difference between each term is 4. Therefore d = 4.

There are n terms.

Substitute all that into your formula for the sum of an arithmetic series and then do with the result what I said to do in my first reply.

**Viral** · October 17th 2009, 02:58 PM

$S_n = \frac{n}{2}[2a + (n-1)d]$

Just to check, is this the formula for calculating the sum?

**mr fantastic** · October 17th 2009, 03:00 PM

Originally Posted by Viral

$S_n = \frac{n}{2}[2a + (n-1)d]$

Just to check, is this the formula for calculating the sum?

It's just as easy for you to go to a textbook or use Google as it is for any of us ....

**Viral** · October 17th 2009, 03:01 PM

It's a help forum after all >> . My textbook has been misplaced, hence why I'm here. I'm also here because google brought me here.

**mr fantastic** · October 17th 2009, 03:07 PM

Originally Posted by Viral

It's a help forum after all >> . My textbook has been misplaced, hence why I'm here. I'm also here because google brought me here.

Yes, help as in help you to help yourself. Really, why should we be the ones that check a formula using Google when you can easily do it yourself? That's your job, not our job.

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