Answer to Question #321361 in Discrete Mathematics for Rakesh

Question

Prove that for every positive integer ‘n’ : 1𝑋2𝑋3 + 2𝑋3𝑋4 + ⋯ + 𝑛(𝑛 + 1)(𝑛 + 2) =

𝑛(𝑛+1)(𝑛+2)(𝑛+3)

Answer 1

Let $P(n)$ be the proposition that 

$$1\times 2\times 3+2\times 3\times 4+...+n(n+1)(n+2)$$

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

Basis step: $P(1)$ is true, because

$$1\times 2\times 3=6=\dfrac{1(1+1)(1+2)(1+3)}{4}$$

Inductive step: For the inductive hypothesis we assume that $P(k)$ holds for an arbitrary positive integer $k.$ That is, we assume that

$$1\times 2\times 3+2\times 3\times 4+...+k(k+1)(k+2)$$

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

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

$$1\times 2\times 3+2\times 3\times 4+...+k(k+1)(k+2)$$

$$+(k+1)(k+1+1)(k+1+2)$$

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

is also true.

We obtain

$$1\times 2\times 3+2\times 3\times 4+...+k(k+1)(k+2)$$

$$+(k+1)(k+1+1)(k+1+2)$$

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

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

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

This last 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

$$1\times 2\times 3+2\times 3\times 4+...+k(k+1)(k+2)$$

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

for all positive integers $n.$