The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Fp1 Series

Q) Show that 1^3 3^3 +5^3 +...+ (2n-1)^3 can be written as the (sum of r^3 from r=1 to 2n) subtract (the sum of (2r)^3 from r=1 to n)
(It's best if you write this out yourself using sigma notation)
I don't know exactly what the best way to do it is but should I show that
the sum of (2r-1)^3 from r=1 to n is equal to something in terms of n [using 0.25n^2*(n+1)^2]... then show that this is equal to the underlined stuff above in terms of n?
Thanks!
Original post by MathMeister
Q) Show that 1^3 3^3 +5^3 +...+ (2n-1)^3 can be written as the (sum of r^3 from r=1 to 2n) subtract (the sum of (2r)^3 from r=1 to n)
(It's best if you write this out yourself using sigma notation)
I don't know exactly what the best way to do it is but should I show that
the sum of (2r-1)^3 from r=1 to n is equal to something in terms of n [using 0.25n^2*(n+1)^2]... then show that this is equal to the underlined stuff above in terms of n?
Thanks!


I'm not sure how much working they are looking for , but it would probably be enough to just say something like:

13+33+⋯+(2n−1)3=13+23+33+⋯+(2n)3−(23+43+⋯+(2n)3)=∑r=12nr3−∑r=1n(2r)31^3+3^3+\dots+(2n-1)^3 = 1^3+2^3+3^3+\dots+(2n)^3 - (2^3+4^3+\dots+(2n)^3) = \displaystyle\sum_{r=1}^{2n}r^3 - \displaystyle\sum_{r=1}^n (2r)^3.
Reply 2
Avatar for ztibor
ztibor
10
Original post by MathMeister
Q) Show that 1^3 3^3 +5^3 +...+ (2n-1)^3 can be written as the (sum of r^3 from r=1 to 2n) subtract (the sum of (2r)^3 from r=1 to n)
(It's best if you write this out yourself using sigma notation)
I don't know exactly what the best way to do it is but should I show that
the sum of (2r-1)^3 from r=1 to n is equal to something in terms of n [using 0.25n^2*(n+1)^2]... then show that this is equal to the underlined stuff above in terms of n?
Thanks!


I think the best way would be the using of induction

AS brittanna wrote You have to show that

∑r=12nr3−∑r=1n(2r)3=∑r=1n(2r−1)3 \displaystyle\sum_{r=1}^{2n}r^3 - \sum_{r=1}^n (2r)^3= \sum_{r=1}^n (2r-1)^3

For n=1

(13+23)−(2⋅1)3=9−8=1=(2⋅1−1)3=13=1 \displaystyle (1^3+2^3)-(2\cdot 1)^3=9-8=1 = (2 \cdot 1-1)^3=1^3=1

Let the equation be true for n

show that will be true for n+1
Original post by ztibor
I think the best way would be the using of induction

AS brittanna wrote You have to show that

∑r=12nr3−∑r=1n(2r)3=∑r=1n(2r−1)3 \displaystyle\sum_{r=1}^{2n}r^3 - \sum_{r=1}^n (2r)^3= \sum_{r=1}^n (2r-1)^3

For n=1

(13+23)−(2⋅1)3=9−8=1=(2⋅1−1)3=13=1 \displaystyle (1^3+2^3)-(2\cdot 1)^3=9-8=1 = (2 \cdot 1-1)^3=1^3=1

Let the equation be true for n

show that will be true for n+1

I wouldn't use induction, personally. I'd just write what brittanna did: "look, it's obvious, that's what it means to be a sum".

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you