27. Calculate the sum

$$
1 \cdot 2+2 \cdot 3+3 \cdot 4+\ldots+n(n+1)
$$

An arithmetic progression is a sequence in which $u_{n+1}=u_{n}+d$ for all $n$. The number $d$ is called the common difference of the progression.

A geometric progression is a sequence in which $u_{n+1}=u_{n} \cdot q$ for all $n$. The number $q$ is called the common ratio of the progression.
27. Let $u_{n}=1 \cdot 2+2 \cdot 3+\ldots+n(n+1)$.

Then $\Delta u_{n}=n(n+1)$ is a polynomial of the second degree. Therefore, from the result of problem 24, it follows that $u_{n}$ is a polynomial of the 3rd degree. Let

$$
u_{n}=a_{0} n^{3}+a_{1} n^{2}+a_{2} n+a_{3}
$$

To find $a_{0}, a_{1}, a_{2}, a_{3}$, substitute the values $n=0,1,2,3$. We get a system of four equations with 4 unknowns:

$$
\left.\begin{array}{l}
u_{0}=a_{3}=0 \\
u_{1}=a_{0}+a_{1}+a_{2}+a_{3}=2 \\
u_{2}=8 a_{0}+4 a_{1}+2 a_{2}+a_{3}=8 \\
u_{3}=27 a_{0}+9 a_{1}+3 a_{2}+a_{3}=20
\end{array}\right\}
$$

which reduces to the system

$$
\left.\begin{array}{r}
a_{0}+a_{1}+a_{2}=2 \\
8 a_{0}+4 a_{1}+2 a_{2}=8 \\
27 a_{0}+9 a_{1}+3 a_{2}=20
\end{array}\right\}
$$

Subtract the first equation from the second and three times the first from the third. We get

$$
\left.\begin{array}{r}
3 a_{0}+a_{1}=2 \\
24 a_{0}+6 a_{1}=14
\end{array}\right\}
$$

In the resulting system, subtract six times the first equation from the second:

$$
6 a_{0}=2 ; a_{0}=\frac{1}{3} ; a_{1}=1 ; a_{2}=\frac{2}{3}
$$

Therefore,

$$
\begin{aligned}
u_{n}=\frac{1}{3} & n^{3}+n^{2}+\frac{2}{3} n= \\
& =\frac{n^{3}+3 n^{2}+2 n}{3}=\frac{n(n+1)(n+2)}{3}
\end{aligned}
$$

i.e.

$1 \cdot 2+2 \cdot 3+\ldots+n(n+1)=\frac{n(n+1)(n+2)}{3}$.