Find this sum, in terms of $n$:
I have some hand-written hints from someone else, but can't read his writing:
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Hint: $2^3+4^3+\cdots+(2j)^3=8(1^3+2^3+\cdots+j^3)$.
If you know the sum of the first so many cubes formula, you can use the above to sort out the even cube sum from the odd cube sum.
paw88789
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Considering some concrete examples may help you understand what he does.
If $n=4$, then $$\begin{align}1^3-2^3+3^3-4^3&=(1^3+2^3+3^3+4^3)-2(2^3+4^3)\\&=(1^3+2^3+3^3+4^3)-2\cdot 2^3(1^3+\color{red}{2}^3)\\&=\left(\frac{4\cdot 5}{2}\right)^2-2^4\cdot \left(\frac{2\cdot 3}{2}\right)^2.\end{align}$$ Here, note that $\color{red}{2}$ is $n/2$.
If $n=5$, then $$\begin{align}1^3-2^3+3^3-4^3+5^3&=(1^3+2^3+3^3+4^3+5^3)-2(2^3+4^3)\\&=(1^3+2^3+3^3+4^3+5^3)-2\cdot 2^3(1^3+\color{blue}{2}^3)\\&=\left(\frac{5\cdot 6}{2}\right)^2-2^4\cdot \left(\frac{2\cdot 3}{2}\right)^2.\end{align}$$ Here, note that $\color{blue}{2}$ is $(n-1)/2$.
mathlove
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