Proof by induction: $(1+\alpha)^n\ge 1+n\alpha +\frac{n(n-1)}{2}\alpha ^2$ so I have this problem. It asks me to prove an expression by induction.

Let $n$ be a positive integer, and $\alpha$ any nonnegative real
number. Prove by induction that$$(1+\alpha)^n\ge 1+n\alpha
 +\frac{n(n-1)}{2}\alpha ^2$$

So I let $G(n) = (1+\alpha)^n\ge 1+n\alpha +\frac{n(n-1)}{2}\alpha ^2$. I showed the initial case when $n=0$ and got $1=1$. I then assume $G(k)$ is true $G(k) = (1+\alpha)^k\ge 1+k\alpha +\frac{k(k-1)}{2}\alpha ^2$. I then try to prove $G(k+1)$ is also true. $G(n) = (1+\alpha)^{k+1}\ge 1+(k+1)\alpha +\frac{(k+1)k}{2}\alpha ^2$. However, this is where I am stuck. I'm not sure where to go from here. Thanks.
 A: You have 
$$\begin{align}
(1+\alpha)^{k+1}&=(1+\alpha)(1+\alpha)^k\\\\
&\ge (1+\alpha)\left(1+k\alpha+\frac{k(k-1)}{2}\alpha^2\right)\\\\
&=1+(k+1)\alpha+\frac{k(k+1)}{2}\alpha^2+\frac{k(k-1)}{2}\alpha^3\\\\
&\ge 1+(k+1)\alpha+\frac{k(k+1)}{2}\alpha^2
\end{align}$$
since $\alpha\ge 0$ and $k\ge 1$.
A: Here's a proof by induction
showing that
the inequality is true
for any fixed size initial part
of the binomial theorem.
Theorem:
For fixed $m$
and $n \ge m$,
if $x \ge 0$ then
$(1+x)^n
\ge \sum_{k=0}^m x^k\binom{n}{k}
$.
Initial case:
Expand
$(1+x)^m$
manually.
You will get
$(1+x)^m
=\sum_{k=0}^m x^k\binom{n}{k}
$.
Suppose true for $n$.
Then
$\begin{array}\\
(1+x)^{n+1}
&=(1+x)(1+x)^{n}\\
&\ge(1+x)\sum_{k=0}^m x^k\binom{n}{k}\\
&=\sum_{k=0}^m x^k\binom{n}{k}+\sum_{k=0}^m x^{k+1}\binom{n}{k}\\
&=\sum_{k=0}^m x^k\binom{n}{k}+\sum_{k=1}^{m+1} x^{k}\binom{n}{k-1}\\
&=1+\sum_{k=1}^m x^k(\binom{n}{k}+\binom{n}{k-1})+x^{m+1}\binom{n}{m}\\
&=1+\sum_{k=1}^m x^k\binom{n+1}{k}+x^{m+1}\binom{n}{m}\\
&=\sum_{k=0}^m x^k\binom{n+1}{k}+x^{m+1}\binom{n}{m}\\
&\ge\sum_{k=0}^m x^k\binom{n+1}{k}\\
\end{array}
$
