## [ Sums of numerical sequences and difference series ] Induction (etc.).

$$
1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}
$$
Induction on $m$ easily proves that $1^{3}+2^{3}+3^{3}+\ldots+m^{3}=\left(\frac{m(m+1)}{2}\right)^{2}$. Indeed, the base case of induction

is obvious, so we only need to verify the equality

$$
\frac{m^{2}(m+1)^{2}}{2}+(m+1)^{3}=\frac{(m+1)^{2}(m+2)^{2}}{2}
$$

After canceling out $m+1$ and multiplying by 4, we get the obvious equality $m^{2}+4(m+1)=(m+2)^{2}$. Thus, $1^{3}+2^{3}+3^{3}+\ldots+(2 n-1)^{3}+(2 n)^{3}=\left(\frac{2 n(2 n+1)}{2}\right)$, i.e., $1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}+2^{3}\left(1^{3}+2^{3}+\ldots+n^{3}\right)$ $=\left(\frac{2 n(2 n+1)}{2}\right)^{2}$. Transform the last equality using the fact that $1^{3}+2^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$. As a result, we get $1^{3}+3^{3}+5^{3}+\ldots+(2 n-1)^{3}=n^{2}\left(2 n^{2}-1\right)$.

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