Question #151698
Prove that (1 * 2) + (2 * 3) + (3 * 4) + (4 * 5) + ....+ n (n + 1) = n (n + 1) (n + 2) / 3 for n positive integers
1
Expert's answer
2020-12-17T19:16:56-0500

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

for n=1\text{for }n=1

n(n+1)=12=2n(n+1)= 1*2 =2

n(n+1)(n+2)3=1233=2\frac{n(n+1)(n+2)}{3}=\frac{1*2*3}{3}=2

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

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

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


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

(12)+(23)+....+k(k+1)+(k+1)(k+2)(1 * 2) + (2 * 3) + ....+ k (k + 1)+(k+1)(k+2) =

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

=k(k+1)(k+2)3+(k+1)(k+2)=k(k+1)(k+2)+3(k+1)(k+2)3=(k+1)(k+2)(k+3)3=\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}

i.e\text{i.e}

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

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


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

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

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

therefore, the expression is true for all positive numbers\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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