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FP1: SIGMA notation HELP!

Question 3)a):
http://www.toothill.notts.sch.uk/data/files/dept/maths/aqaalevel/AQA-MFP1-W-QP-JUN05.PDF

I have NO IDEA how to do it! I can do many sigma notation questions on other papers but this one, I don't know.

Please help me! Thank you!!

Reply 1

ghostwalker

Original post by Konnichiwa

...

This may help:

\displaystyle\sum_{i=1}^n [a(b+c)]=\sum_{i=1}^n[ab+ac]=\sum_{i=1}^n ab+\sum_{i=1}^n ac

If you're still stuck, please clarify your problem.

Reply 2

Konnichiwa

I know that, but I am still stuck. I have done so many other similar questions, try it and you will know it's different from the others.

If you can do it, please share with me how you did it.

Reply 3

ghostwalker

Original post by Konnichiwa

I know that, but I am still stuck. I have done so many other similar questions, try it and you will know it's different from the others.

If you can do it, please share with me how you did it.

It's no great difference to any other.

Since you know the bits I mentioned, all you need to is subtract the RHS of one formula from the RHS of the other. Take out a factor of n(n+1)/12 and rearrange the rest.

Reply 4

Konnichiwa

Well it's difficult getting into the form they want, that's what it is different. I've tried doing that and factorising many times, yet no step further.

It is harder than it seems.

Reply 5

ghostwalker

Original post by Konnichiwa

Well it's difficult getting into the form they want, that's what it is different. I've tried doing that and factorising many times, yet no step further.

It is harder than it seems.

It would have help if you'd said as such in your opening post.

Taking out the factors I mentioned gives:

\displaystyle\frac{n(n+1)}{12}[3n(n+1)-2(2n+1)]

Now expand and then factorise what's in the square brackets.

Reply 6

Konnichiwa

Original post by ghostwalker

It would have help if you'd said as such in your opening post.

Taking out the factors I mentioned gives:

\displaystyle\frac{n(n+1)}{12}[3n(n+1)-2(2n+1)]

Now expand and then factorise what's in the square brackets.

Thanks! I finally got it now, I can't believe I did not see that.

Reply 7

ghostwalker

Original post by Konnichiwa

Thanks! I finally got it now, I can't believe I did not see that.

np.

Reply 8

mathsRus

hi I need help on this question please got exam on Friday paper question.png

Reply 9

Phichi

Original post by mathsRus

hi I need help on this question please got exam on Friday paper question.png

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

Does that help?

If you are really stuck, I suggest you go learn the chapter again.

(edited 10 years ago)

Reply 10

mathsRus

O I am sorry I did not specify. I needed help on part b of the question where I did use the answer from previous part but I don't know what put in place of (n)

Thanks for your reply.

Reply 11

ghostwalker

Original post by mathsRus

O I am sorry I did not specify. I needed help on part b of the question where I did use the answer from previous part but I don't know what put in place of (n)

Thanks for your reply.

The smallest odd number after 100 is 101, so what value of n gives 101?

Similarly for largest odd number before 200

You now have the limits on your summation.

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