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Write down the sum $$\sum_{n=1}^{2N}n^3$$ in terms of $N$, and find $1^3-2^3+3^3-4^3+...-(2N)^3$ in terms of $N$, simplifying your answer.

My attempt,

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

I don't know how to solve the next part, can anyone give me some hints? Thank you.

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  • $\begingroup$ $\sum_{n=1}^N (2n)^3 = \sum 8n^3 = 8\sum n^3$ $\endgroup$ – Trevor Norton Jan 7 '16 at 15:20
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Consider the following:

$1^3-2^3+3^3-4^3\pm...+(2N-1)^3-(2N)^3$

$=[1^3+2^3+3^3+...+(2N)^3]-2[2^3+4^3+...+(2N)^3]$

$=[1^3+2^3+3^3+...+(2N)^3]-16[1^3+2^3+...+N^3]$

So, if we define $S(n)=1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4}$, then the sum in consideration is $S(2N)-16S(N)$

Hammering through the algebra, we find:

$S(2N)-8S(N)=\frac{4N^2(2N+1)^2}{4}-16\frac{N^2(N+1)^2}{4}=N^2((2N+1)^2-4(N+1)^2)=N^2(-4N-3)=-N^2(4N+3)$

This method can also be used to compute Dirichlet eta function in terms of the Riemann zeta function.

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Subtract out twice the even terms.

$$\sum_{k=1}^N (-1)^{k+1} k^3 = \sum_{k=1}^{2 N} k^3 - 2 \sum_{k=1}^{N } (2 k)^3 $$

which is

$$\frac14 (2 N)^2 (2 N+1)^2 - \frac14 16 N^2 (N+1)^2 = N^2 (2 N+1)^2 - 4 N^2 (N+1)^2 = - N^2 (4 N+3) $$

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Hint: If you add twice the sum of even terms to your desired quantity, you get $\sum_{n=1}^{2N} n^3$. To compute the sum of even terms, factor out a common factor of $8$ to get $8 \sum_{n=1}^N n^3$.

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You can write $$1^3-2^3+3^3-4^3+ \ldots - (2N)^3$$ $$=1^3+2^3+3^3+4^3+ \ldots + (2N)^3 -2\left[2^3+4^3+6^3+\ldots +(2N)^3\right]$$ $$=1^3+2^3+3^3+4^3+ \ldots + (2N)^3 -2^4\left[1^3+2^3+3^3+\ldots +N^3\right]$$

Using the result you derived,

The above expression $$=\left[\frac{2N(2N+1)}{2}\right]^2-16\left[\frac{N(N+1)}{2}\right]^2$$ $$=N^2(2N+1)^2-4N^2(N+1)^2$$ $$=N^2(4N^2+4N+1-4N^2-8N-4)$$ $$=-N^2(4N+3)$$

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