It is known that
$$\begin{array}{ccc}1+2&=&3 \\ 4+5+6 &=& 7+8 \\ 9+10+11+12 &=& 13+14+15 \\\ 16+17+18+19+20 &=& 21+22+23+24 \\\ 25+26+27+28+29+30 &=& 31+32+33+34+35 \\\ldots&=&\ldots \end{array}$$
There is something similar for square numbers:
$$\begin{array}{ccc}3^2+4^2&=&5^2 \\ 10^2+11^2+12^2 &=& 13^2+14^2 \\ 21^2+22^2+23^2+24^2 &=& 25^2+26^2+27^2 \\ \ldots&=&\ldots \end{array}$$
As such, I wonder if there are similar 'consecutive numbers' for cubic or higher powers. Of course, we know that there is impossible for the following holds (by Fermat's last theorem): $$k^3+(k+1)^3=(k+2)^3 $$