Question

Prove that (1 + x)^n > or equal to (1 + nx) , for all natural number n , where x>-1

Answer 1

@ajprincess @FoolForMath @siddhantsharan @satellite73 @experimentX

Answer 2

Mathamatical Induction!

Answer 3

try induction

Answer 4

not getting!

Answer 5

ok did you do the base step?

Answer 6

yes P(1)

Answer 7

then assume p(k) is true

Answer 8

ok so what do we get to assume? that
\[(1+x)^k\geq 1+kx\] and then we need to show that given this fact, we can conclude that
\[(1+x)^{k+1}\geq 1+(k+1)x\] right?

Answer 9

we didn't do anything yet, i just wanted to make sure this is the track we are taking

Answer 10

ok

Answer 11

the reason these are confusing is that it is not an equality, so it is not clear how to manipulate the expression to get what you want. the gimmick here is to multiply both sides of the inequality we get to assume, namely
\[(1+x)^k\geq 1+kx\] by \(1+x\) and we get to keep the senses of the inequality because \(x>-1\) so \(1+x>0\)

Answer 12

how would you know this step? i don't know the answer, but it is the quickest way to get \((1+x)^{k+1}\) out of \((1+x)^k\)

Answer 13

yes i knw it!!

Answer 14

would you like to try it yourself, or would you like me to continue?

Answer 15

i have tried this and i am unable u plzz go on!

Answer 16

oh ok, so given the induction hypothesis, namely that
\[(1+x)^k\geq 1+kx\] we multiply by \(1+x\) and get
\[(1+x)^k(1+x)\geq (1+kx)(1+x)\]
\[\implies (1+x)^{k+1}\geq 1+x+kx+kx^2\]
\[\implies (1+x)^{k+1}\geq 1+(k+1)x+kx^2\] and since \(x^2>0, k>1\) we know \(kx^2>0\) so
\[(1+x)^{k+1}\geq 1+(k+1)x\] as required

Answer 17

thxxx

Answer 18

yw
notice that the induction took care of itself, all that was necessary was some algebra

Answer 19

yup it needs some logical thinking!!!

Answer 20

Here is a proof for x >0. Use the Binomial Expansion Theorem
\[
(1+x)^n = 1 + n x + {n \choose 2} x^2 + \cdots + x^n \ge 1 + n x
\]