Prove that (1 * 2) + (2 * 3) + (3 * 4) + (4 * 5) + ....+ n (n + 1) = n (n + 1) (n + 2) / 3 for n positive integers

$\text{We use the method of mathematical induction}$

$\text{for }n=1$

$n(n+1)= 1*2 =2$

$\frac{n(n+1)(n+2)}{3}=\frac{1*2*3}{3}=2$

$\text{for }n=1 \text{ statement is true}$

$\text{suppose the statement is true for } n =k$

$(1 * 2) + (2 * 3) + ....+ k (k + 1) = \frac{k (k + 1) (k + 2) }{3}\ (1)$

$\text{сonsider the expression for } n = k+1$

$(1 * 2) + (2 * 3) + ....+ k (k + 1)+(k+1)(k+2)$ =

$\text{from equality (1):}$

$=\frac{k (k + 1) (k + 2) }{3}+(k+1)(k+2)= \frac{k (k + 1) (k + 2)+3(k+1)(k+2) }{3}=\frac{ (k + 1) (k + 2)(k+3) }{3}$

$\text{i.e}$

$(1 * 2) + (2 * 3) + ....+ k (k + 1)+(k+1)(k+2)=$ $\frac{ (k + 1) (k + 2)(k+3) }{3}$

$\text{ the statement is true for } n =k+1$

$\text{the expression is true for n = 1 on the assumption}$

$\text{that the expression is true for n = k follows that}$

$\text{the expression is also true for n= k+1}$

$\text{therefore, the expression is true for all positive numbers}$

Answer:the statement under the conditions of the problem is correct,

proven by mathematical induction