Find the least value of n for which $\displaystyle \sum_{r=1}^n (4r-3) > 2000$
How do I do this?
Furst thing you do is simplify it down by extracting the constants outside:
$\displaystyle \sum_{r=1}^n (4r-3) = 4 \sum_{r=1}^n r - \sum_{r=1}^n 3 = 4 \sum_{r=1}^n r - 3n$.
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.
$\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.
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.
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.