mathematical induction ($(1+x)^n\ge1+nx+n(n-1)x^2/2$) Suppose that $x > 0$ and let $n \geq 2$ be a positive integer. Prove that $(1 + x)^n \geq 1 + nx + \frac{n(n-1)}{2}x^2$
So for the base case, I have $x=1$, but that really is not getting me anywhere.
Would it be better to manipulate $n$?  Is there a way to know which one to manipulate?
 A: Why does anyone need mathematical induction here? Just use the Binomial theorem:
$$
(1+x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = 1 + \binom{n}{1}x + \binom{n}{2} x^2 + R(n,x)
$$
where $R(n,x)$ is strictly positive because $x>0, n>0$ hence the statement. 
A: HINT:
Let $(1+x)^m\ge \{1+mx+\frac{m(m-1)}2x^2\}$
So, $(1+x)^{m+1}=(1+x)(1+x)^m \ge (1+x)\{1+mx+\frac{m(m-1)}2x^2\}$
$=1+(m+1)x+x^2\{\frac{m(m-1)}2+m\}+x^3\frac{m(m-1)}2\ge \{1+(m+1)x+\frac{m(m+1)}2x^2\}$  as $x>0$ and $m\ge2$
Now, for $m=2,1+2x+x^2-\{1+2x+x^2\}=0\implies (1+x)^2\ge \{1+2x+\frac{2(2-1)}2x^2\}$
A: Mathematical induction proof proceeds as follows:
1.Step: one checks that the statement (in your case the given inequality) holds for $n=1$.
2.Step: one makes an assumption that the inequality holds for $n$. Then, with that assumption in mind, one should show that it also follows for $n+1$. That is, in your case, you should show that 
$(1+x)^{n+1}\geq 1+(n+1)x+\frac{(n+1)n}{2}x^{2}$
holds provided that
$(1+x)^{n}\geq 1+nx+\frac{n(n-1)}{2}x^{2}$
is true.
Hope this helps!
A: *

*Show that the statement is true for $n=0$

*Suppose that for a given number $n$, $(1+x)^n>1+nx+\frac{x(x+1)}{2}x^2$

*Prove that $(1+x)^{n+1}> 1+(n+1)x+\frac{(n+1)(n+2)}{2}x^2$

*Conclude : The statement is true for $n=0$ and for each n that the statement holds, it holds also for $n+1$. Therefore, $\forall n \in \Bbb N$, $(1+x)^n>1+nx+\frac{x(x+1)}{2}x^2$
