Solving a summation problem - I know the answer I think, but not why it works This problem I intuitively know the answer, (I think) but I am trying to get a handle on why it works.
$$\sum_{n=0}^x n=\frac{\sum_\limits{n=0}^x n^3}{\sum_\limits{n=0}^x n^2} $$
We want to know what x is.
I figured the answer was x = 1, because the only number I can come up with where $\frac{n^3 + (n+1)^3 + ... }{n^2 + (n+1)^2 + ...}=n$ is one, when I did a few sums and realized that anything other than n=1 doesn't work. And I know it's because if I add a number I am not getting a cube over a square of n anymore, $\frac{(1^3 + 2^3)}{(1^2+2^2)}=\frac{9}{4}$ for example. But I feel like there is some basic piece of a proof/procedure I am missing, even though this intuition is correct. Anyhow, I am curious what the "real" reason might be -- I suspect this is some relatively simple algebraic principle that makes this work, I feel like I am just not quite seeing it. Or maybe I got it right and just overthought this.
 A: These are well-known sums: $$\sum_{0}^x n = \frac{x(x+1)}{2} \\ \sum_0^x n^2 = \frac{x(x+1)(2x+1)}{6} \\ \sum_0^x n^3 = \frac{x^2(x+1)^2}{4} $$ Substituting in gives $$\frac{x(x+1)}{2} = \frac{\frac{x^2(x+1)^2}{4}}{\frac{x(x+1)(2x+1)}{6}} \\ \implies \frac{2x+1}{3} =1 \implies x=1 $$ is the only solution.
See here for the proofs of these identities.
A: The simplest proof that $\sum_ii\sum_jj^2=\sum_kk^3$ requires (zero terms or) one term only in each sum is to note the cases with $i=j$ already give us every term on the right-hand side, so we need to avoid positive cross terms $ij^2$ with $i\ne j$.
A: The sum of cube is the square of the triangular numbers, viz $(n^2+n)^2/4$,
So the equation is correct if you drop the square from the $n^2$ im the denominator.
A: A more elementary way:  Multiply both sides by the denominator to get
$$(1+2+\cdots+x)(1^2+2^2+\cdots+x^2) = (1^3+2^3+\cdots+x^3).$$
See that on the left, if you mupltiply everything out, you get every cube plus some other junk.  If you subtract the cubes from both sides, you get
$$\mbox{junk} = 0$$
which means you'd better not have any "cross terms" which means that $x$ had better be $1.$
A: Let's say the initial equation is: $$ S_1 = \frac{S_2}{S_3} $$
It is well known that:
$$S_1 = \frac{x(x+1)}{2}$$
$$S_2 = {\frac{x^2(x+1)^2}{4}}$$
$$S_3 = \frac{x(x+1)(2x+1)}{6}$$
Plug these into the original equation and solve for $x$.
