Question

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

Answer 1

Let P(n) = $1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\cdots +\frac{1}{(1+2+3+\cdots +n)}=\frac{2n}{(n+1)}$
For n = 1
$P\left(1\right)=1=\frac{2\times 1}{1+1}\Rightarrow 1=1\therefore P\left(1\right)$ is true
Let P (n0 be true for n = k
$\therefore P\left(k\right)=1+\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+\cdots +\frac{1}{(1+2+3+\cdots +k)}=\frac{2k}{k+1}Forn=k+1P(k+1)=-1+\frac{1}{(1+2)}+\frac{1}{1+2+3}+\cdots +\frac{1}{(1+2+3+\cdots +k)}+\frac{1}{(1+2+3+\cdots +k+1)}=\frac{2k}{k+1}+\frac{1}{(1+2+3+\cdots +k+1)}$
$\text{Now}1+2+3+\cdots +k+1\text{Is an A.P.}\therefore 1+2+3+\cdots +k+1=\frac{(k+1)}{2}[1+k+1]=\frac{(k+1)(k+2)}{2}\therefore P(k+1)=\frac{2k}{k+1}+\frac{1}{\frac{(k+1)(k+2)}{2}}=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{2}{k+1}[k+\frac{1}{k+2}]=\frac{2}{k+1}\left[\frac{k^2+2k+1}{k+2}\right]=\frac{2}{k+1}\left[\frac{(k+1{)}^2}{k+2}\right]=\frac{2(k+1)}{(k+2)}$
$\therefore$ P (k + 1) is true
Thus P (k) is true $\Rightarrow$ P (k + 1) is true
Hence by principle fo mathematical induction,
P(n) is true for all $nϵN.$