The Student Room Group

FP1 question

Can someone explain why this

Screen Shot 2016-05-19 at 15.56.53.pngequals this:

(with (2r-1) replacing ....)

I'm so confused, any help is appreciated!
Reply 1
Since you are starting from the nth term plus one and going to the 2*nth term. the sum is equal to the sum from 1 to the 2*nth term, taking away the sum of the terms before the nth term plus one. Which is the sum from one to n of 2r-1. I hope that makes sense although I will draw out an example and post it afterwards.
Reply 2
Original post by wamble
Can someone explain why this

Screen Shot 2016-05-19 at 15.56.53.pngequals this:

(with (2r-1) replacing ....)

I'm so confused, any help is appreciated!


Think about the definition of sigma.

For example, let's say I wanted to sum r=36ar=a3+a4+a5+a6\displaystyle \sum_{r=3}^{6} a_r = a_3 + a_4 + a_5 + a_6.

We have: r=16ar=a1+a2+a3+a4+a5+a6\displaystyle \sum_{r=1}^{6} a_r = a_1 + a_2 + a_3 + a_4 + a_5 + a_6

And: r=12ar=a1+a2\displaystyle \sum_{r=1}^2 a_r = a_1 + a_2

Subtracting those two:r=16arr=12ar=(a1+a2+a3+a4+a5+a6)(a1+a2)=a3+a4+a5+a6=r=36ar\displaystyle \sum_{r=1}^{6} a_r - \sum_{r=1}^2 a_r = (a_1 + a_2 + a_3 + a_4 + a_5 + a_6) - (a_1 + a_2) = a_3 + a_4 + a_5 + a_6 = \sum_{r=3}^6 a_r

So, as you can see, if you want to sum between two number r=n+12nar\sum_{r=n+1}^{2n} a_r, you can instead sum from 11 to 2n2n and to get a1++a2na_1 + \cdots + a_{2n} and then subtract away the first 1 to (n1)(n-1) terms to get: an+1++a2na_{n+1} + \cdots + a_{2n} as required.

Hence: r=n+12nar=r=12narr=1nar\displaystyle \sum_{r=n+1}^{2n} a_r = \sum_{r=1}^{2n} a_r - \sum_{r=1}^{n} a_r
Reply 3
It just saying that the sum from (say) 500 to (say) 900 is equal to the sum from 1 to 900 take away the sum from 1 to 499.
This question is common and the error usually made is by one term, in the middle. e.g using 500 instead of 499.

Where the algebra is easier it sometimes make sense to do (in the context of my example) 1 to 900 then take off 1 to 500 and then add back in the 500th term
Reply 4
Since Zacken has already replied I won't send the follow up. I hope his explanation helps.
Reply 5
Original post by spico
Since Zacken has already replied I won't send the follow up. I hope his explanation helps.


:cool:
Reply 6
haha thanks guys, that makes a lot of sense :smile:
Reply 7
Original post by wamble
haha thanks guys, that makes a lot of sense :smile:


No problem. :-)

Quick Reply

Latest