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`,
` L.H.S. = 1` and `R.H.S. = (2 xx 1)/(1+1) = 2/2 = 1`
Thus, ` P(1)` is true.
Let `P(n)` be true for some `n = k`.
i.e, `1+(1)/(1+2)+"……"+(1)/(1+2+3)+"......"+(1)/(1+2+3+"...."k) = (2k)/(k+1)"....."(1)`
Now, we have 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+2+3+"...."+(k+1))`
` = (2k)/(k+1)+(1)/(1+2+3+"....."+(k+1))`
`= (2k)/(k+1)+(1)/(((k+1)(k+2))/(2))`
[Using `1+2+3+"...."+n = (n(n+1))/(2)`]
`= (2k)/((k+1)) + (2)/((k+1)(k+2))`
`= (2)/(k+1)((k(k+2)+1)/(k+2))`
`= (2(k+1)^(2))/((k+1)(k+2))`
`= (2(k+1))/((k+2))`
Thus, `P(k+1)` is the whenever `P(k)` is true.
Hence by the principle of mathemetical induction, statement `P(n)` is true for all natural numbers.