Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\cdots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e,
$P\left(n\right):\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\cdots +\frac{1}{n(n+1)(n+2)}$
$=\frac{n(n+3)}{4(n+1)(n+2)}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):\frac{1}{1\cdot 2\cdot 3}=\frac{1(1+3)}{4(1+1)(1+2)}$
$\Rightarrow \frac{1}{6}=\frac{4}{4\cdot 2\cdot 3}$
$\Rightarrow \frac{1}{6}=\frac{1}{6}$
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,
$P\left(K\right):\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\cdots +\frac{1}{K(K+1)(K+2)}$
$=\frac{K(K+3)}{4(K+1)(K+2)}\cdots \left(1\right)$

Step $\left(4\right):$ Checking $P\left(n\right)$ for $n=K+1$
Now we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\cdots +\frac{1}{K(K+1)(K+2)}+\frac{1}{(K+1)(K+2)(K+3)}$
$=\frac{K(K+3)}{4(K+1)(K+2)}+\frac{1}{(K+1)(K+2)(K+3)}$ (using $\left(1\right)$)
$=\frac{1}{(K+1)(K+2)}[\frac{K(K+3)}{4}+\frac{1}{(K+3)}]$
$=\frac{1}{(K+1)(K+2)}\left[\frac{K(K+3{)}^2+4}{4(K+3)}\right]$
$=\frac{1}{(K+1)(K+2)}\left[\frac{K(K^2+6K+9)+4}{4(K+3)}\right]$
$=\frac{1}{(K+1)(K+2)}\left[\frac{K^3+6K^2+9K+4}{4(K+3)}\right]$

We can write it as

$=\frac{1}{(K+1)(K+2)}\left[\frac{K^3+2K^2+K+4K^2+8K+4}{4(K+3)}\right]$
= $\frac{1}{(K+1)(K+2)}\left[\frac{K(K^2+2K+1)+4(K^2+2K+1)}{4(K+3)}\right]$
$=\frac{1}{(K+1)(K+2)}\left[\frac{(K^2+2K+1)(K+4)}{4(K+3)}\right]$
$=\frac{(K+1{)}^2(K+4)}{4(K+1)(K+2)(K+3)}$
$=\frac{(K+1)(K+4)}{4(K+2)(K+3)}$
We can write it as
= $\frac{(K+1)\left(\right(K+1)+3)}{4\left(\right(K+1)+1)\left(\right(K+1)+2)}$
Thus $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final answer:
Therefore, by the principle of mathematical induction statement $P\left(n\right)$ is true for all $n\in N$.