Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$1\cdot 2\cdot 3+2\cdot 3\cdot 4+\cdots +n(n+1)(n+2)$
$=\frac{n(n+1)(n+2)(n+3)}{4}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.,
$P\left(n\right):1\cdot 2\cdot 3+2\cdot 3\cdot 4+\cdots +n(n+1)(n+2)=\frac{n(n+1)(n+2)(n+3)}{4}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1\cdot 2\cdot 3=\frac{1(1+1)(1+2)(1+3)}{4}$
$\Rightarrow 6=\frac{1\cdot 2\cdot 3\cdot 4}{4}$
$\Rightarrow 6=6$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i. e.,
$P\left(K\right):1\cdot 2\cdot 3+2\cdot 3\cdot 4+\cdots +K(K+1)(K+2)=\frac{K(K+1)(K+2)(K+3)}{4}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$1\cdot 2\cdot 3+2\cdot 3\cdot 4+\cdots +K(K+1)(K+2)+(K+1)(K+2)(K+3)$
$=\frac{K(K+1)(K+2)(K+3)}{4}+(K+1)(K+2)(K+3)$
$=(K+1)(K+2)(K+3)(\frac{K}{4}+1)$
$=\frac{(K+1)(K+2)(K+3)(K+4)}{4}$
$=\frac{(K+1)\left[\right(K+1)+1]\left[\right(K+1)+2]\left[\right(K+1)+3]}{4}$
Thus $P(K+1)$ is true, whenever $P\left(K\right)$ is true.
Final Answer:
Hence, from the principle of mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.