Thread: induction question

I have to show the following by mathematical induction:

for each n in the natural numbers and for a > -1, the following inequality holds
(1 + a)^n >= 1 + na + (n/2)(n-1)a^2 + (n/6)(n-1)(n-2)a^3.

I know how to do the base case and the induction step.

For n + 1, the left hand side of the inequality is (1+a)^(n+1) = (1+a)(^n)*(1+a). Then I multiplied the right hand side of the inequality by (1+a) assuming that the "n" case is correct.

After multiplying through, I get the following:
1 + (n+1)a + (n/2)(n+1)a^2 + (n/6)(n+1)(n-1)a^3 + (n/6)(n-1)(n-2)a^4. The first four terms are precisely the original right hand side of the inequality in "n+1." (i.e. what I'm looking for). However, how do I know the left hand side is still greater than or equal to the right hand side with the additional term (n/6)(n-1)(n-2)a^3 under these conditions. Do you know what I'm saying? I think I did all the arithmetic and algebra correctly.

Thanks for any help or suggestions.

Originally Posted by PvtBillPilgrim

I have to show the following by mathematical induction:

for each n in the natural numbers and for a > -1, the following inequality holds
(1 + a)^n >= 1 + na + (n/2)(n-1)a^2 + (n/6)(n-1)(n-2)a^3.

I know how to do the base case and the induction step.

For n + 1, the left hand side of the inequality is (1+a)^(n+1) = (1+a)(^n)*(1+a). Then I multiplied the right hand side of the inequality by (1+a) assuming that the "n" case is correct.

After multiplying through, I get the following:
1 + (n+1)a + (n/2)(n+1)a^2 + (n/6)(n+1)(n-1)a^3 + (n/6)(n-1)(n-2)a^4. The first four terms are precisely the original right hand side of the inequality in "n+1." (i.e. what I'm looking for). However, how do I know the left hand side is still greater than or equal to the right hand side with the additional term (n/6)(n-1)(n-2)a^3 under these conditions. Do you know what I'm saying? I think I did all the arithmetic and algebra correctly.

Thanks for any help or suggestions.

It looks vaguely like a binomial expansion of some sort. Are you sure that it isn't ? If that was the case then your result follows easily. Else if we have a large enough positive a then

for any .

-Dan

It is related to the Bernoulli inequality. Maybe that page will help.