Proof By Induction (difficult problem)

[Dish]

Hey guys, I'm just totally stumped by this Q

Use mathematical induction to prove that for integer n, n > or = to 2,
(x+1)^n - nx - 1 is divisible by x^2.

It's not a Homework problem.

Attempt at solution:

1) Prove for n =2,
(x+1)^2 - 2x - 1 = x^2 + 2x +1 - 2x -1 = x^2
2) Assume true for n = k
thus (x+1)^k -kx -1 = m x^2 where m is any integer
3) Prove for n = k+1

(x+1)^(k+1) - (k+1)x - 1
(x+1)^k . (x+1) - kx - x - 1
(x+1)^k - kx -1 + x(x+1)^k - x - 1
m.x^2 + x(x+1)^k - x -1

from here on i am totally confused. :S
Please can someone help me to finish the proof, so that it's divisible by x^2?
Thank you :)

 

[Mute]

Hey guys, I'm just totally stumped by this Q

Use mathematical induction to prove that for integer n, n > or = to 2,
(x+1)^n - nx - 1 is divisible by x^2.

It's not a Homework problem.

Attempt at solution:

1) Prove for n =2,
(x+1)^2 - 2x - 1 = x^2 + 2x +1 - 2x -1 = x^2
2) Assume true for n = k
thus (x+1)^k -kx -1 = m x^2 where m is any integer
3) Prove for n = k+1

(x+1)^(k+1) - (k+1)x - 1
(x+1)^k . (x+1) - kx - x - 1
(x+1)^k - kx -1 + x(x+1)^k - x - 1
m.x^2 + x(x+1)^k - x -1

At the bold line, you can factor out (x+1) from some of the terms so that you have $(x+1)(\mbox{something})+(\mbox{something else})$. Look at the "something" carefully & think how you can relate it to your kth step. Plugging that relation in for the "something", you should be able to then expand things and get the result of x^2*(stuff).

By the way, your assumption that at the kth step (x+1)^k -kx - 1 = mx^2, where m is an integer, is incorrect. m can be a polynomial in x.

 

[Dish]

argh Thanks I get it now, thanks for the help