numbers' pattern 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):
$$k^3+(k+1)^3=(k+2)^3 $$
 A: We do have
$$ \eqalign{6^3 &= 3^3 + 4^3 + 5^3\cr
            20^3 &= 11^3 + 12^3 + 13^3 + 14^3\cr
            40^3 &= 3^3 + \ldots + 22^3\cr
            70^3 &= 15^3 + \ldots + 34^3\cr
37^3 + 38^3 &= 5^3 + \ldots + 25^3\cr
  30^3 + 31^3 + 32^3 &= 7^3 + \ldots + 24^3\cr 
            101^3 + 102^3 + 103^3 &= 61^3 + \ldots + 71^3\cr
 15^3 + \ldots + 20^3 &= 11^3 + \ldots + 19^3\cr
            681^3 + \ldots + 687^3 &= 566^3 + \ldots + 577^3\cr
 \cr}$$
and many others.
EDIT: See also http://oeis.org/A062682
A: There is the famous $3^3+4^3+5^3=6^3$. But I don't know if there are others.
At least not where the whole sequence is consecutive...
One gets the problem of three triangular numbers T1 T2 T3 for which one wants to solve $T1^2+T2^2=2 \cdot T3^2$.
A: Let's start by proving the basic sequences and then where and why trying to step it up to cubes fails. I don't prove anything just reduce the problem to a two variable quartic Diophantine equation.

Lemma $1 + 2 + 3 + 4 + \ldots + n = T_1(n) = \frac{n(n+1)}{2}$.
Corollary $(k+1) + (k+2) + \ldots + (k+n) = -T_1(k) + T_1(k+n)$
The first sequence of identities is $$-T_1(s(n)) + T_1(s(n)+n+1) = -T_1(s(n)+n+1) + T_1(s(n)+2n+1)$$ so computing
? f(x) = (x*(x+1))/2
? (-f(s)+f(s+n+1))-(-f(s+n+1)+f(s+2*n+1))
% = -n^2 + (s + 1)

we find $s(n) = n^2-1$ and prove it.

Lemma $1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = T_2(n) = \frac{n(n+1)(2n+1)}{6}$.
The second sequence of identities is $$-T_2(s(n)) + T_2(s(n)+n+1) = -T_2(s(n)+n+1) + T_2(s(n)+2n+1)$$ so computing
? f(x) = (x*(x+1)*(2*x+1))/6
? (-f(s-1)+f(s-1+n+1))-(-f(s-1+n+1)+f(s-1+2*n+1))
% = -2*n^3 + (-2*s - 1)*n^2 + s^2

this is a weird quadratic equation in two integers with some solutions (n,s) = (1,3), (2,10), (3,21), (4,36), (5,55), (6,76), ...
the discriminant of the polynomial (as a polynomial in $s$) is $2^2 n^2 (n+1)^2$
 so actually we can solve it and that explains where there's one solution for each $n$.

Now lets try for cubes.. but at this point we know it's not going to work
? f(x) = ((x^2+x)/2)^2
? (-f(s-1)+f(s-1+n+1))-(-f(s-1+n+1)+f(s-1+2*n+1))
% = -7/2*n^4 + (-6*s - 3)*n^3 + (-3*s^2 - 3*s - 1/2)*n^2 + s^3

so this is too complicated to actually solve but if anyone proves this doesn't have solutions for positive $n$ that will show there are no such cubic sequences.
For reference $$7n^4 + (12s + 6)n^3 + (6s^2 + 6s + 1)n^2 - 2s^3 = 0$$ is the Diophantine equation that obstructs a cubic sequence from existing. 
Maybe you could conclude by the Mordell Conjecture that there's no infinite family of sequences of identities for cubic and higher power sums, if you can show these polynomials are always irreducible.
