Problem 961

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.

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