Using the principle of mathematical induction, prove each of the following

Solution:

To Prove:

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

Let us prove this question by principle of mathematical induction (PMI)

Let $P(n): \frac{1}{1}+\frac{1}{(1+2)}+\frac{1}{(1+2+3+\ldots \ldots+n)}=\frac{2 n}{(n+1)}$

For n = 1

$\mathrm{LHS}=1$

$\mathrm{RHS}=\frac{2 \times 1}{(1+1)}=1$

Hence, LHS = RHS

$P(n)$ is true for $n=1$

Assume $P(k)$ is true

$\frac{1}{1}+\frac{1}{(1+2)}+\ldots \ldots+\frac{1}{(1+2+3+\ldots \ldots \times+k)}=\frac{2 k}{(k+1)}$ ……(1)

We will prove that P(k + 1) is true

$\mathrm{RHS}=\frac{2(k+1)}{(k+1+1)}=\frac{2 k+2}{k+2}$

$\mathrm{LHS}=\frac{1}{1}+\frac{1}{(1+2)}+\ldots \ldots+\frac{1}{(1+2+3+\ldots \ldots+(k+1))}$

$=\frac{1}{1}+\frac{1}{(1+2)}+\ldots \ldots+\frac{1}{(1+2+3+\ldots \ldots+k)}+\frac{1}{(1+2+3+\ldots \ldots+(k+1))}$ [Writing the last Second term]

$=\frac{2 k}{(k+1)}+\frac{1}{(1+2+3+\ldots \ldots+(k+1))}$ [From 1]

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

$\{1+2+3+4+\ldots+n=[n(n+1)] / 2$ put $n=k+1\}$

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

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

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

[Taking LCM and simplifying]

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

= RHS

Therefore, ${ }^{\frac{1}{1}}+\frac{1}{(1+2)}+\ldots \ldots+\frac{1}{(1+2+3+\ldots \ldots \times(k+1))}=\frac{2 k+2}{k+2}$

LHS = RHS

Therefore, $P(k+1)$ is true whenever $P(k)$ is true.

By the principle of mathematical induction, $P(n)$ is true for $x$

Where $\mathrm{n}$ is a natural number

Put k = n - 1

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

Hence proved