Using the principle of mathematical induction prove that `1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)++1/(1+2+3++n)=(2n)/(n+1)` for all `n in N`

Explanation

`P(n):1+1/(1+2)+1/(1+2+3+)….+1/(1+2+3+….+n)=2n/(n+1)`

For `n=1,(2×1)/(1+1)=2/2=1
`Thus, P(1) is true.

Let P(n) be true for some n=k.
To prove that P(n) is true for `n=k+1. ie.,
1+1/(1+2)+...+1/(1+2+3+)...+1/(1+2+3+.....+(k+1)) =(2(k+1))/(k+2) `

Adding `1/(1+2+3+......+(k+1)) `on both sides of (1), we get `1+1/(1+2+)....+1/(1+2+3+)....1/(1+2+3+....+k+1)+1/(1+2+3+....+(k+1))`

`=(2k)/(k+1)+1/(1+2+3+.....+(k+1)) `

`=(2k)/(k+1)+2/((k+1)(k+2))) `

`=(2k)/(k+1)+2/((k+1)(k+2)) `

`=(2)/(k+1)(k(k+2)+1)/(k+2) `

...