Question

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

Answer 1

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

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

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
$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\cdots +\frac{1}{(1+2+...+K)}+\frac{1}{(1+2+\cdots +(K+1\left)\right)}$
$=\frac{2K}{K+1}+\frac{1}{(1+2+3+\cdots +(K+1\left)\right)}$ (Using $\left(1\right)$)

$\because$ We know that
$1+2+3+\cdots +n=\frac{n(n+1)}{2}$
So, we get
$=\frac{2K}{K+1}+\frac{1}{\frac{(K+1)(K+1+1)}{2}}$
$=\frac{2K}{K+1}+\frac{2}{(K+1)(K+2)}$
$=\frac{2k(K+1)+2}{(K+1)(K+2)}$
$=\frac{2}{(K+1)(K+2)}[K^2+2K+1]$
$=\frac{2(K+1{)}^2}{(K+1)(K+2)}$
= $\frac{2(K+1)}{(K+2)}$
We can write it as, $=\frac{2(K+1)}{\left[\right(K+1)+1]}$
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 $n\in N$.