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

How do I do this?

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- Oct 17th 2009, 02:33 PMViralFind the least value of n in an arithmetic sequence
Find the least value of n for which $\displaystyle \sum_{r=1}^n (4r-3) > 2000$

How do I do this? - Oct 17th 2009, 02:40 PMMatt Westwood
Furst thing you do is simplify it down by extracting the constants outside:

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

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

Then you have an inequality in n to solve. Good luck. - Oct 17th 2009, 02:42 PMViral
Hmm, I'm not really sure what you did there =\ . We've never covered that.

- Oct 17th 2009, 02:44 PMmr fantastic
$\displaystyle \sum_{r=1}^n (4r-3) = 4 \left( \sum_{r=1}^n r\right) - 3n $ and you should know the formula to substitute for $\displaystyle \sum_{r=1}^n r$. 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. - Oct 17th 2009, 02:44 PMMacstersUndead
- Oct 17th 2009, 02:46 PMViral
Is $\displaystyle \sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))$?

- Oct 17th 2009, 02:49 PMMacstersUndead
- Oct 17th 2009, 02:51 PMViral
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.

- Oct 17th 2009, 02:52 PMMatt Westwood
- Oct 17th 2009, 02:56 PMViral
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.

- Oct 17th 2009, 02:56 PMmr fantastic
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. - Oct 17th 2009, 02:58 PMViral
$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d]$

Just to check, is this the formula for calculating the sum? - Oct 17th 2009, 03:00 PMmr fantastic
- Oct 17th 2009, 03:01 PMViral
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.

- Oct 17th 2009, 03:07 PMmr fantastic