Put them all in

$\begin{aligned} \displaystyle \sum^{10}_{n=1} n(1+n+n^2) & = \displaystyle \sum^{10}_{n=1}n+n^2+n^3 \\ & = \frac{10\cdot 11}{2}+\frac{10\cdot 11 \cdot 21}{6}+\left( \frac{10\cdot 11}{2}\right)^{2} \\ & = 55+385+55^2 \\ & = \boxed{3465} \end{aligned}$

$\text{Formulae used:} \\ \displaystyle \sum^{k}_{n=1}n=\frac{k(k+1)}{2} \\ \displaystyle \sum^{k}_{n=1}n^2=\frac{k(k+1)(2k+1)}{6} \\ \displaystyle \sum^{k}_{n=1}n^3=\left(\frac{k(k+1)}{2}\right)^2$

If you know the formulae for the sums of n, n squared and n cubed, you can do it quite quickly.

Considering that the the sum of n(1+n+n^2) can be expanded to n+n^2+n^3, we can apply the following formulas:

Sum of n= (n (n+1))/2 Sum of n squares (n^3)/3 + (n^2)/2 + n/6 Sum of n cubes= ((n (n+1))/2)^2

We substitute: (10x11)/2 + 1000/3 + 100/2 + 10/6 + ((10x11)/2)^2

Which equals to 55 + 385 + 3025 = 3465

You rewrite the sum as: (k(k+1)/2)(1+(2k+1)/3 + k(k+1)/2), where k = 10 => 55(1+7+55) (= 55(60+3) = 165 + 3300)=3465 The steps in parentheses are for not doing any computer calculation.

$=\sum_{n=1}^{10}n+\sum_{n=1}^{10}n^2+\sum_{n=1}^{10}n^3=\frac{10\cdot11}{2}+\frac{10\cdot11\cdot(2 \cdot 10+1)}{6}+\left(\frac{10\cdot 11}{2}\right)^2=55+385+55^2=440+(5\cdot6\cdot100+25)=440+3025=3465$

10×11/2+{(10)×(10+1)×(10×2+1)}/6+(10^2)(10+1)^2/4 =55+385+3025 =3465