The Student Room Group

A2 C4 Maths help - Binomial expansion question

I've been going through the 2015 C4 maths paper and cannot figure out how you do the second part to this question:
Find the binomial expansion of up to and including the term in . [3 marks]

I think the answer to this is: 1/4 - x/16 +( 5x^2)/256

(ii) Use your expansion from part (b)(i) to find an estimate for , giving your answer to four decimal places. [2 marks]

Any help on (ii) would be really appreciated!
Thank you.
Reply 1
Original post by EmJ15
I've been going through the 2015 C4 maths paper and cannot figure out how you do the second part to this question:
Find the binomial expansion of up to and including the term in . [3 marks]

I think the answer to this is: 1/4 - x/16 +( 5x^2)/256

(ii) Use your expansion from part (b)(i) to find an estimate for , giving your answer to four decimal places. [2 marks]

Any help on (ii) would be really appreciated!
Thank you.


what you do is use trial and improvement and try and make your become , by playing around with the first equation once that x value gives you the second equation what you do is sub in that value of x into your expansion and find the value :smile:
Reply 2
So the expansion is valid for x<8/3 x<8/3 . What happens if you let x=1/3.
Reply 3
Original post by EmJ15
I've been going through the 2015 C4 maths paper and cannot figure out how you do the second part to this question:
Find the binomial expansion of up to and including the term in . [3 marks]

I think the answer to this is: 1/4 - x/16 +( 5x^2)/256


This bit is correct.

(ii) Use your expansion from part (b)(i) to find an estimate for , giving your answer to four decimal places. [2 marks]

Any help on (ii) would be really appreciated!
Thank you.


If you notice that:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\left(\frac{1}{81}\right)^{1/3} = \left(\frac{1}{9^2}\right)^{1/3} = (9^{-2})^{1/3} = 9^{-2/3}\end{equation*}



Then it is fairly obvious that x=1/3x = 1/3 gives you (8+1)2/3(8 + 1)^{-2/3} and you can plug this into your binomial expansion to get an approximation.
Reply 4
Original post by Zacken
This bit is correct.



If you notice that:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\left(\frac{1}{81}\right)^{1/3} = \left(\frac{1}{9^2}\right)^{1/3} = (9^{-2})^{1/3} = 9^{-2/3}\end{equation*}



Then it is fairly obvious that x=1/3x = 1/3 gives you (8+1)2/3(8 + 1)^{-2/3} and you can plug this into your binomial expansion to get an approximation.


Thank you! That makes more sense now. :smile:
Reply 5
Original post by EmJ15
Thank you! That makes more sense now. :smile:


You're welcome! :smile:

Quick Reply