Question

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

Answer 1

Restrict x > -1, so 1+x > 0.
- For n = 1, obviously (1+x)1 = (1+x) = 1 + nx.


- Suppose true for n.
Then (1+x)n+1 = (1+x)n(1+x)
>= (1+nx)(1+x) since we're supposing the statement true for n and since (1+x) > 0
= 1 + nx + x + nx2 by multiplication
>= 1 + nx + x since x2 >= 0
= 1 + (n+1)x QED.

Note that substituting for a multiplicative quantity in inequalities only works when all elements are positive, so we need (1+x) > 0

Answer 2

hey it is Mathamatecial Induction Problem

Answer 3

@ujjwal @UnkleRhaukus @Callisto @myininaya @heena @eseidl @quarkine

Answer 4

(1+x)^n>1+nx

Answer 5

((1+x)^n)(1+x)>(1+nx)(1+x)
(1+x)^n+1>(1+nx+x+x^2n)>1+(n+1)x.........

Answer 6

how????

Answer 7

see (1+x)^n>1+nx............u assume this r8

Answer 8

yup!

Answer 9

then multiply both side with (1+x)

Answer 10

since x>-1 sign doesnt change

Answer 11

OK

Answer 12

mathlover's answer is induction. looks ok except for the lousy syntax, which makes it hard to decipher.

Answer 13

yea...mathlover already gave the answer.!!

Answer 14

but i did nt understand plzzz @A.Avinash_Goutham plzz continoue

Answer 15

soo LHS is (1+x)^(n+1)............