Question

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

Answer 1

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

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

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
$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +K(K+1)+(K+1)(K+2)$
$=\frac{K(K+1)(K+2)}{3}+(K+1)(K+2)$ (Using $\left(1\right)$)
$=(K+1)(K+2)(\frac{K}{3}+1)$
$=\frac{(K+1)(K+2)(K+3)}{3}$
We can write it as,
$=\frac{(K+1)\left\{\right(K+1)+1\}\left\{\right(K+1)+2\}}{3}$
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 $n\in N$.