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FP1 Series Question... Need help please..

I can do part a and part b correctly, but can't find do part C :frown:

Really need help please!!!
I can use the answer from part B but I can't see the connection between A and C and can't get the correct answer :frown:

Reply 1

RDKGames

Study Forum Helper

Original post by Oliviazh

I can do part a and part b correctly, but can't find do part C :frown:

Really need help please!!!
I can use the answer from part B but I can't see the connection between A and C and can't get the correct answer :frown:

Well you could just expand the quadratic and do 3 different sums. Can't go wrong with that.

(edited 7 years ago)

Reply 2

the bear

i don't really think you need to use part a) for part c)

:redface:

Reply 3

RDKGames

Study Forum Helper

Original post by the bear

i don't really think you need to use part a) for part c)

:redface:

Just an off-topic question; why do you always put asterisks on everything you write?? lol

Reply 4

Oliviazh

Original post by RDKGames

Well you could just expand the quadratic and do 3 different sums. Can't go wrong with that.

Well I did that but still wrong answer...must be something wrong with the calculation :frown:

Reply 5

RDKGames

Study Forum Helper

Original post by Oliviazh

Well I did that but still wrong answer...must be something wrong with the calculation :frown:

Why are you using a calculator?? I mean, you're allowed yeah, but for what? There's nothing to calculate. You're probably factoring wrong. Show your working if you get stuck

(2r-1)^2=4r^2-4r+1

\displaystyle \sum_{r=1}^{n} (4r^2-4r+1) = \sum_{r=1}^n (4r^2) + \sum_{r=1}^{n} (-4r) + \sum_{r=1}^{n} (1)

The first term you can say that

4r^2=(2r)^2

and use the result from part b.

(edited 7 years ago)

Reply 6

Oliviazh

Original post by RDKGames

Why are you using a calculator?? I mean, you're allowed yeah, but for what? There's nothing to calculate.

(2r-1)^2=4r^2-4r+1

\displaystyle \sum_{r=1}^{n} (4r^2-4r+1) = \sum_{r=1}^n (4r^2) + \sum_{r=1}^{n} (-4r) + \sum_{r=1}^{n} (1)

The first term you can say that

4r^2=(2r)^2

and use the result above.

Um..I didn't use a calculator...I mean there must be something wrong with my calculation😂
Just a quick question, for the last term in your equation, does the result of that part equal to 1 or n?

Reply 7

RDKGames

Study Forum Helper

Original post by Oliviazh

Um..I didn't use a calculator...I mean there must be something wrong with my calculation😂
Just a quick question, for the last term in your equation, does the result of that part equal to 1 or n?

Misread but there's still no calculation anyway lol. And yes. You are summing up 1 and

n

amount of times so it's 1*n hence n.

(edited 7 years ago)

Reply 8

Oliviazh

Original post by RDKGames

Yes. You are summing up 1 and

n

amount of times so it's 1*n hence n.

Thank you😃

Reply 9

Zacken

Meh, the above is overkill.

Better to think of it as summing the odd terms up till 2n-1, you know the sum of all terns from 1 to 2n and you know the sum of even terms from 1 to 2n. So the sum of all terms - sum of even terms (i.e: part a - part b) gets you what you want.