Army of GOD wrote:Haggis_McMutton wrote:Army of GOD wrote:Question whatever: Prove that (1+x)^n ≤ 1+nx where x > -1 and n is any integer that is greater than 1.
Wait. For x=1, n=3 => [(1+1)^3 <= 1+1*3] <=> [2^3 <= 4] ... ?
whoops, that's supposed to be greater than or equal to, not less than.
Allrighty then.
Proof by induction on n that: [(1+x)^n >= 1+nx] for any x in (-1, inf), n in N*.
P(1) : 1+x >= 1+x (True)
P(n): (1+x)^n >= 1+nx
P(n+1): (1+x)^(n+1) >= 1+(n+1)x <=>
(1+x)^(n+1) >= 1+ nx + x
Notice that if (1+x)^(n+1) >= (1+x)^n + x, then by P(n) (1+x)^(n+1) >= 1+nx+x so If i prove (1+x)^(n+1) >= (1+x)^n + x i get the result
x is in (-1, inf) so 1+x > 0 so i divide by (1+x)^n =>
1+x >= 1 + x/(1+x)^n <=>
x >= x/(1+x)^n
Now we have 2 cases:
1. x is in [0, inf) => 1 >= 1/(1+x)^n <=>
(1+x)^n >= 1
(1+x) is in [1, inf) so that's true for all n in N*
2. x is in (-1,0) => 1 <= 1/(1+x)^n <=>
(1+x)^n <= 1
(1+x) is now in (0,1) so that's again true for all n in N*
so P(n) implies P(n+1) for all n in N*