Series? What series?

Assuming $x$ to be an integer, this equation can be rewritten as

$\displaystyle\sum_{k=1}^{x} k^{3} + \sum_{k=1}^{x} k = 110$

$\Longrightarrow \left(\dfrac{x(x + 1)}{2}\right)^{2} + \dfrac{x(x + 1)}{2} = 110.$

Now let $y = \dfrac{x(x + 1)}{2}.$ Then our equation becomes

$y^{2} + y - 110 = 0 \Longrightarrow (y + 11)(y - 10) = 0.$

Now clearly $y \gt 0,$ so we must have that

$y = \dfrac{x(x + 1)}{2} = 10 \Longrightarrow x^{2} + x - 20 = 0 \Longrightarrow (x + 5)(x - 4) = 0.$

Since clearly $x \gt 0$ we can conclude that $x = \boxed{4}.$

given,

2+10 + 3*10+...+x(x^2+1)=110

42+...+x(x^2+1)=110

therefore,

110-42=68

4(4^2+1)=68

therefore, x=4