Question

Check whether the statement is true or false then true give reason

$1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+...+\frac{1}{(1+2+3+...n)}=\frac{2n}{(n+1)}$

• A. True
• B. False

Answer 1

The correct option is A True

Let $P\left(n\right)$ be the given statement, then by principle of mathematical induction,

$P\left(n\right)=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +n}=\frac{2n}{n+1}$

Let $n=1$,

$P\left(1\right)=\frac{2\times 1}{1+1}$

$=1$

So, the statement is true.

Let $n=k$,

$P\left(k\right)=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +k}=\frac{2k}{k+1}$

It is to prove that $P\left(k+1\right)$ is true for $n=k+1$.

Consider,

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +k}+\frac{1}{1+2+3+\cdots +k+\left(k+1\right)}$

$=\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +k}\right)+\frac{1}{1+2+3+\cdots +k+\left(k+1\right)}$

$=\frac{2k}{k+1}+\frac{1}{1+2+3+\cdots +k+\left(k+1\right)}$

$=\frac{2k}{k+1}+\frac{1}{\frac{\left(k+1\right)\left(k+1+1\right)}{2}}$ $\left(\sum n=\frac{n\left(n+1\right)}{2}\right)$

$=\frac{2k}{k+1}+\frac{2}{\left(k+1\right)\left(k+1+1\right)}$

$=\frac{2k}{k+1}\left(k+\frac{1}{k+2}\right)$

$=\frac{2k}{k+1}\left(\frac{k^2+2k+1}{k+2}\right)$

$=\frac{2{\left(k+1\right)}^2}{\left(k+1\right)\left(k+2\right)}$

$=\frac{2\left(k+1\right)}{k+2}$

$=\frac{2n}{n+1}$

This shows that $P\left(k+1\right)$ is true when $P\left(k\right)$ is true.

Hence, $P\left(n\right)$ is true for all natural numbers.