Answer to Question #225272 in Discrete Mathematics for Zeeshan

Question

Prove that for every positive integer n. 1.2.3+2.3.4++ n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4.

Answer 1

Let $P(n)$ be the proposition that for every positive integer $n,$

$$1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$$

BASIC STEP: $P(1)$ is true, because $1\cdot2\cdot3=\dfrac{1(1+1)(1+2)(1+3)}{4}$

INDUCTIVE STEP: For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary

positive integer $k.$ That is, we assume that

$$1\cdot2\cdot3+2\cdot3\cdot4+...+k(k+1)(k+2)=\dfrac{k(k+1)(k+2)(k+3)}{4}$$

Under this assumption, it must be shown that $P(k+1)$ is true, namely, that

$$1\cdot2\cdot3+2\cdot3\cdot4+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)=$$

$$=\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}$$

is also true.

$$1\cdot2\cdot3+2\cdot3\cdot4+...+k(k+1)(k+2)+(k+1)(k+2)(k+3)$$

$$=\dfrac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)$$

$$=\dfrac{(k+1)(k+2)(k+3)(k+4)}{4}$$

This equation shows that $P(k + 1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we

know that $P(n)$ is true for all positive integers $n.$ That is, we have proven that

$$1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}$$

for all positive integers $n.$