# Sums of Series?

• Sep 17th 2011, 10:09 AM
Sums of Series?
Given that

S(N)=

N
Σ...(2n-1)^3
n=1

Show that
S(N)=N^2(2N^2 - 1)

I tried expanding it and used the following formulas:
Σn=n(n+1)/2
Σn^2=n(n+1)(2n+1)/6
Σn^3=n^2(n+1)^2/4

But it doesn't give the answer.

I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

Thanks
• Sep 17th 2011, 11:56 AM
chisigma
Re: Sums of Series?
Quote:

Given that

S(N)=

N
Σ...(2n-1)^3
n=1

Show that
S(N)=N^2(2N^2 - 1)

I tried expanding it and used the following formulas:
Σn=n(n+1)/2
Σn^2=n(n+1)(2n+1)/6
Σn^3=n^2(n+1)^2/4

But it doesn't give the answer.

I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

Thanks

Starting with the general formula...

$\displaystyle S_{n}= \sum_{k=1}^{n} k^{3}= \frac{n^{2}\ (n+1)^{2}}{4}$ (1)

... we derive from (1)...

$\displaystyle S_{2n-1}= \sum_{k=1}^{2n-1} k^{3}= n^{2}\ (2n-1)^{2}$ (2)

$\displaystyle S_{2n}= \sum_{k=1}^{n} (2k)^{3}= 2 n^{2} (n+1)^{2}$ (3)

.. so that is...

$\displaystyle \sum_{k=1}^{n} (2k-1)^{3} = S_{2n-1} - S_{2n}$ (4)

The last details are left to You...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Sep 27th 2011, 03:47 AM
Re: Sums of Series?
Quote:

Given that

S(N)=

N
Σ...(2n-1)^3
n=1

Show that
S(N)=N^2(2N^2 - 1)

I tried expanding it and used the following formulas:
Σn=n(n+1)/2
Σn^2=n(n+1)(2n+1)/6
Σn^3=n^2(n+1)^2/4

But it doesn't give the answer.

I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

Thanks

$\displaystyle \sum_{n=1}^ N n^3=\frac{N^2(N+1)^2}{4}$

$\displaystyle \sum_{n=1}^N (2n-1)^3=1^3+3^3+5^3+.......+(2N-1)^3$

$\displaystyle =1^3+2^3+3^3+....+(2N-2)^3+(2N-1)^3-\left[2^3+4^3+....(2N-2)^3\right]$

$\displaystyle =\sum_{n=1}^{2N-1} n^3-\left[2^3+4^3+6^3+....+(2N-2)^3\right]$

$\displaystyle =\sum_{n=1}^{2N-1} n^3-2^3\left[1^3+2^3+3^3+....+(N-1)^3\right]$

$\displaystyle =\sum_{n=1}^{2N-1} n^3-2^3\sum_{n=1}^{N-1} n^3$

and now you may use your formula for the sum of cubes.
• Sep 27th 2011, 05:10 AM
CaptainBlack
Re: Sums of Series?
Quote:

Given that

S(N)=

N
Σ...(2n-1)^3
n=1

Show that
S(N)=N^2(2N^2 - 1)

I tried expanding it and used the following formulas:
Σn=n(n+1)/2
Σn^2=n(n+1)(2n+1)/6
Σn^3=n^2(n+1)^2/4

But it doesn't give the answer.

I also tried substituting (2n-1) directly into the last formula but it also doesn't give the answer.

Thanks

Have you considered using induction?

CB
• Sep 27th 2011, 06:09 AM
Soroban
Re: Sums of Series?

Quote:

$\displaystyle \text{Given that: }\:S(N) \;=\;\sum^N_{n=1}(2n-1)^3$

$\displaystyle \text{Show that: }\;S(N)\:=\:N^2(2N^2 - 1)$

I tried expanding it and used the following formulas:

. . $\displaystyle \sum n \:=\:\frac{n(n+1)}{2}$

. . $\displaystyle \sum n^2 \:=\:\frac{n(n+1)(2n+1)}{6}$

. . $\displaystyle \sum n^3 \:=\:\frac{n^2(n+1)^2}{4}$

But it doesn't give the answer. .
It should!

$\displaystyle \sum(2n-1)^3 \;=\;\sum(8n^3 - 12n^2 + 6n - 1)$

. . . . . . . . . $\displaystyle =\;8\left(\sum n^3\right) - 12\left(\sum n^2\right) + 6\left(\sum n\right) - \left(\sum 1\right)$

. . . . . . . . . $\displaystyle =\;8\cdot\frac{N^2(N+1)^2}{4} - 12\cdot\frac{N(N+1)(2N+1)}{6} + 6\cdot\frac{N(N+1)}{2} - N$

. . . . . . . . . $\displaystyle =\;2N^2(N+1)^2 - 2N(N+1)(2N+1) + 3N(N+1) - N$

. . . . . . . . . $\displaystyle =\;2N^2(N^2+2N+1) - 2N(2N^2 + 3N + 1) + 3N^2 + 3N - N$

. . . . . . . . . $\displaystyle =\;2N^4 + 4N^3 + 2N^2 - 4N^3 - 6N^2 - 2N + 3N^2 + 3N - N$

. . . . . . . . . $\displaystyle =\;2N^4 - N^2$

. . . . . . . . . $\displaystyle =\;N^2(2N^2-1)$