"\\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{\u0441onsider 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
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