Use mathematical induciton to prove that the formula is true for all natural numbers p.?

Here the Problem 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/p*(p+1) = p/p+1Use mathematical induciton to prove that the formula is true for all natural numbers p.?
When p = 1,


1/1*2 = 1/2 and


p/(p+1) = 1/(1+1) = 1/2, so


1/1*2 = p(p+1)








Assume that


1/1*2 + 1/2*3 + 1/3*4 + ... + 1/p*(p+1) = p/p+1 is true.


Then


1/1*2 + 1/2*3 + 1/3*4 + ... + 1/p*(p+1) + 1/(p+1)*(p+2)


= p/(p+1) + 1/(p+1)*(p+2) .... (By adding 1/(p+1)*(p+2) to both sides of the original equation. From here down, I am simplifying this side of the equation, while leave the LHS alone.)


= (p+2)p/(p+2)*(p+1) + 1/(p+1)*(p+2)


= (p^2 + 2p + 1)/(p+2)*(p+1)


= (p+1)(p+1)/(p+2)*(p+1)


= (p+1)/(p+2)


This means that, if it is true for p, then it is true for p+1.


I already showed that it is true when p=1, so now we know that it is true when p=2, which means that it is also true when p=3, p=4, p=5, p=6, and so on.





So, by induction,


1/1*2 + 1/2*3 + 1/3*4 + ... + 1/p*(p+1) = p/(p+1) is true for all counting number values of p.Use mathematical induciton to prove that the formula is true for all natural numbers p.?
when p = 1, 1/2 = 1/2, trivial


when p = 2, 1/2 + 1/6 = 3/6 + 1/6 = 4/6 = 2/3


just to get a better feel of it, when p = 3,


1/2 + 1/6 + 1/12 = 6/12 + 2/12 + 1/12 = 9/12 = 3/4





so assume the formula is true up to p-1. we show it's true for p.


1/2 + 1/6 + ... + 1/(p-1)p + 1/p(p+1) =


(p-1)/p + 1/p(p+1) =


(p+1)(p-1)/p(p+1) + 1/p(p+1) =


(p虏 - 1)/p(p+1) + 1/p(p+1) =


p虏/p(p+1) =


p/p+1