AS Level further maths series question

Original post by Amy.fallowfield

Hi, is anyone able to explain this for me. The marks schemes get to 2 equations with different values for n and r and I'm not sure why.

Show that the sum of the cubes of the first n positive odd numbers is
n^2 (2n^2 − 1)

file:///C:/Users/amyry/Downloads/Ch.3%20Series.pdf
(Questions on page 1 as Q1 and mark scheme on page 4)

The expression for the sum of the first n cubes is one you can quote.

You need to find

1^3 + 3^3 + 5^3 + ... + (2n-1)^3

This is the same as the sum of all the integers cubed up to (2n-1)^3 minus the sum of the even integers cubed i.e.

= (1 ^3 + 2^3 + 3^3 + ... + (2n-1)^3) - (2^3 + 4^3 + 6^3 + ... + (2n-2)^3)

The first bracket is the series you can quote (up to 2n-1 instead of n). For the second one try to take out a common factor and post your working if you get stuck.

(edited 9 months ago)

Reply 2

Ignorabimus

9

In the formula book, iirc, you are given all the necessary summations required here, it's just a demonstrating your understanding question.
I'll denote the sum from a=0 to n as S(), where a is an arbitrary variable.
So from the formula book or memory:

S(r), integers, is n(n+1)/2
S(r^2), squares, is n(n+1)(2n+1)/6
S(r^3), cubes, is [n(n+1)/2]^2, also commonly known as the square of S(r).

Then notice that you can let A be in the set of naturals with 0.
So, 2A+1 is always odd for all A in the set.
Then S(2A+1) is the sum of odds and so S([2A+1]^3) is the sum of odds cubed.
Expand [2A+1]^3 and you can separate the different powers of A into separate summations and take out the coefficients to the front of the summations.
Since you know what S(r), S(r^2) and S(r^3) are, A is likewise similar and equal in the context of this, so you can substitute the summations for their equivalent formulae and finally simplify your answer.

Reply 3

OP

Thank you I get it now :smile:

I also got to n=6 for part b but dont understand where the answer 11 comes from

The sum of the cubes of 10 consecutive positive odd numbers is 99 800
(b) Use the answer to part (a) to determine the smallest of these 10 consecutive positive
odd numbers.

Reply 4

Original post by Amy.fallowfield

Thank you I get it now :smile:

I also got to n=6 for part b but dont understand where the answer 11 comes from

The sum of the cubes of 10 consecutive positive odd numbers is 99 800
(b) Use the answer to part (a) to determine the smallest of these 10 consecutive positive
odd numbers.

So you have

\displaystyle \sum_{r=k}^{r=k+9} (2r-1)^3 = 99800

And that sum can be split like this

\displaystyle \sum_{r=k}^{r=k+9} (2r-1)^3 = \sum_{r=1}^{r=k+9} (2r-1)^3 - \sum_{r=1}^{r=k-1} (2r-1)^3

Does that help?

Reply 5

OP

Original post by Notnek

So you have

\displaystyle \sum_{r=k}^{r=k+9} (2r-1)^3 = 99800

And that sum can be split like this

\displaystyle \sum_{r=k}^{r=k+9} (2r-1)^3 = \sum_{r=1}^{r=k+9} (2r-1)^3 - \sum_{r=1}^{r=k-1} (2r-1)^3

Does that help?

I did that and substituted the equation from part a in, thats how I got to 6, I just don't get how you get from 6 to 11

Reply 6

Original post by Amy.fallowfield

I did that and substituted the equation from part a in, thats how I got to 6, I just don't get how you get from 6 to 11

Please post your working.

Reply 7

OP

https://drive.google.com/file/d/1KPj8p1SfmrAMUwfHbmtwIaWya4N09HC1/view?usp=sharing

Original post by Notnek

Please post your working.

Reply 8

Original post by Amy.fallowfield

I did that and substituted the equation from part a in, thats how I got to 6, I just don't get how you get from 6 to 11

Ah you would have got k = 6 using what I wrote above which is correct. But it asks you for the smallest of these positive integers so you need to sub k = 6 into 2k-1. The "6" represents the 6th odd number.

(edited 9 months ago)

Reply 9

OP

Original post by Notnek

Ah you would have got k = 6 using what I wrote above which is correct. But it asks you for the smallest of these positive integers so you need to sub k = 6 into 2k-1.

OHHHHH that makes sense. Thank you so much! :smile:

Reply 10

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