Prove by Mathematical Induction Given any real number x if 1

Prove by Mathematical Induction:

Given any real number x, if 1+x>0, then (1+x)^n>=1+n*x

Solution

Denote the given statement: if (1+x) > 0 then (1+x)^n >=1+n*x, by P(n).

First of all, it may be observed that the statement P(1) i.e., (1+x) > 0 => (1+x)^n >=1+n*x holds good, as

                                                                                   (1+x)^1= (1+x) = 1+1*x

Now assume that P(k) i.e., (1+x) > 0 => (1+x)^k >= 1+k*x ...(1)

For n=k+1, we have (1+x) ^ (k+1)= { (1+x) ^ k} *(1+x)

From relation (1), substituting (1+x)^k=1+k*x into the last equation, we get:

(1+x) ^ (k+1)= (1+k*x)*(1+x) ....(2)

Also, since (1+x)>0, it follows from (2) that (1+x) ^ (k+1) >= (1+k*x),

i.e., the given relation holds for n=1, and if it is assumed that it holds for n=k, it holds also for n=k+1, i.e., P(k+1) is true.