PLease help with simple c1 question

Find the least value of n for which :

n
å (4r-3) > 2000
r = 1

å = epsilon sign for sum of.

Anyone? Any help will be greatly aprpeciated. Thanks!

Reply 1

Christophicus

slop8

Find the least value of n for which :

n
å (4r-3) > 2000
r = 1

å = epsilon sign for sum of.

Anyone? Any help will be greatly aprpeciated. Thanks!

n/2(2 + 4(n-1)) > 2000
n/2(2 + 4n - 4) > 2000
n/2(4n - 2) > 2000
2n^2 - n > 2000
2n^2 - n - 2000 > 0

Then solve the inequality.

Reply 2

Irrational Pi

Just tried this question (not OP)...
It doesn't factor, so use quadratic eq.

--1 +- sqrt(1) - (4x2x-2000)
therefore, -1 +- 1 - 16,000...

I get either -15,998/4, which probably isn't right for C1, as you'd really need a calculator.

Or... -16,000/4, which comes out to quite "neatly" be -4000.

So... n>4000?

Unless I've made a mistake, which is all together too likely.

Reply 3

Fermat

You don't need to use a calculator to solve a quadratic, you can just look at the inequality and make a guess.

You have,

2n² - n > 2000

forget the n term, since it is small compared to the n² term. This gives you,

2n² > 2000
n² > 1000
n > 10√(10)

and √(10) is about 3, so,

n > 30 (approx)

So take n=30 as the first guess,

n   2n² - n
===========
30  1770
31  1891
32  2016

So answer is n = 32
===============

Latest

Trending

Articles for you

How to write an excellent personal statement in 10 steps

How to write your law personal statement...and make it really stand out

Can you speak lawyer?

Discussion time: what laws protect women’s rights in the UK?

Try out the app

Continue on web