Jump to content
Sign in to follow this  

May 08 Paper 3 Series And Differential Equations TZ 2 Question 5 b

Recommended Posts

Hi,

Since I've recently switched to my 3rd math teacher in the course of my IB (!), my entire class is lagging behind quite a bit on the option. I've started looking through some past papers and its going surprisingly well so far, but anyway: I've come across a question from the 2008 paper:

me735jyujuew.png

Now my teacher was nice enough to let me take a look at the mark scheme which is along the lines of:

∛(n+1)-n=n(∛(1+1/n3)-1)

=n(1+1/3n3-1/9n6+5/81n9-...-1)

First question is how they got to this last part?

Going on from here, they used v=1/n2 as the "auxillary series". Could someone please explain what this means and what exactly its used for?

I still have problems with some of the later parts of the solution but I think I might be able to figure these out once I get the earlier parts.

Thanks :)

Good luck to everyone else who's sitting the may session!

post-40304-0-39600200-1333433769_thumb.p

Edited by iBS2012

Share this post


Link to post
Share on other sites

a. using ratio test, make the (n+1)/n term less than 1. You should be able to obtain the answer

b. using integral test you should be getter the result

I believe the integration in part b is a bit complicated....

sry my respond maybe a bit superficial... PM me if u need the detail

Share this post


Link to post
Share on other sites

I remember the mark scheme saying that you can only get 3 out of 7 marks for using the integral test.

And check your inbox (:

just checked MS.... I admit my answer is not correct.... seems like MS is using approximation i.e. taylor expansion. to solve this problem....

I bet I won't be able to solve this in the exam...luckily my option is stat but I do have a large opportunity to see such question in FM paper

Share this post


Link to post
Share on other sites

Yea, that was my thought, but how they expanded that series is honestly beyond me, especially with the mark allocation where it is.

Anyone else have any insights on this question?

Share this post


Link to post
Share on other sites

Yea, that was my thought, but how they expanded that series is honestly beyond me, especially with the mark allocation where it is.

Anyone else have any insights on this question?

That is just a normal expansion which you may find it in the textbook... actually what surprised me is the mentioning of the approximation rather than approximation itself

Share this post


Link to post
Share on other sites

If you take the limit of the quotient of two series as n approaches infinity, and the limit is finite and positive, then both series either converge or diverge.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.