Solve the Sum Equation

The equation can be rewritten in sigma notation as $\sum\limits_{k=1}^x (k - 1)k(k + 1) = 7292700$ which simplifies to $\sum\limits_{k=1}^x (k^3 - k) = 7292700$ or $\sum\limits_{k=1}^x k^3 - \sum\limits_{k=1}^x k = 7292700$ . Since the sum of cubes is $\sum\limits_{k=1}^x k^3 = (\frac{x(x+1)}{2})^2$ and the sum of integers is $\sum\limits_{k=1}^x k = \frac{x(x+1)}{2}$ , the equation becomes $(\frac{x(x+1)}{2})^2 - \frac{x(x+1)}{2} = 7292700$ .

Letting $u = \frac{x(x+1)}{2}$ and substituting gives $u^2 - u = 7292700$ and rearranging gives $u^2 - u - 7292700 = 0$ .

Solving this quadratic equation for $u$ gives $u = \frac{ - (-1) \pm \sqrt {(-1)^2 - 4(1)(-7292700)} }{2(1)}$ which simplifies to $u = 2701$ or $u = -2700$ . Since $x > 0$ according to the original equation’s pattern, $u = \frac{x(x+1)}{2} > 0$ , which means $u \neq -2700$ and so $u = 2701$ .

Since $u = \frac{x(x+1)}{2} = 2701$ , $x(x + 1) = 5402$ , $x^2 + x = 5402$ , and $x^2 + x - 5402 = 0$ . Solving this quadratic equation for $x$ gives $x = \frac{ - 1 \pm \sqrt {1^2 - 4(1)(-5402)} }{2(1)}$ which simplifies to $x = 73$ or $x = -74$ . Since $x > 0$ according to the original equation’s pattern, $x \neq -74$ and so $x = 73$ .