Find $n$ for a $\sum_{x=1}^n \left[(x+1)^3-x^3\right]$

Question

$$ \sum_{x=1}^n \left[(x+1)^3-x^3\right]$$ This is my sum, I tried simplfifying and got $3x^2+3x+1$ but Im stuck on how to resolve the sum for $n$.

Answer 1

Note that

$$\sum_{x=1}^n \left[(x+1)^3-x^3\right] =\color{red}{2^3}-1^2+\color{red}{3^3}-\color{red}{2^3}+\ldots+\color{red}{n^3}-\color{red}{(n-1)^3}+(n+1)^3-\color{red}{n^3}$$

Answer 2

As you stated: $$ \sum_{x=1}^{n} [(x+1)^3-x^3] = \sum_{x=1}^{n} 3x^2+3x+1 = 3\frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n = $$ $$ = n(n+1)(n+2)+n = (n+1)^3-1$$ Keep in mind that: $$ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6} $$ $$ \sum_{x=1}^{n} x = \frac{n(n+1)}{2}$$