Prove using the binomial theorem

Question

I have this homework question:

Let $n$ be a positive integer, and $\alpha$ any nonnegative real number. Use Binomial Theorem to show that $$(1+\alpha)^n\ge 1+n\alpha+\frac{n(n-1)}{2}\alpha^2$$

I know that the binomial theorem states that $(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}\cdot y^{k}$. When applying this to the given expression I get $(1+\alpha)^n=\binom{n}{0}+\binom{n}{1}\alpha+\binom{n}{2}\alpha^2+\binom{n}{3}\alpha^3+\cdots +\alpha^n$. I'm not sue how I can show that this is $\ge 1+n\alpha+\frac{n(n-1)}{2}\alpha^2$. Thanks

Answer 1

Since every term is positive, $$ (1+\alpha)^n={n\choose 0}+{n\choose 1}\,\alpha+{n\choose 2}\,\alpha^2+\cdots \\ \geq{n\choose 0}+{n\choose 1}\,\alpha+{n\choose 2}\,\alpha^2 \\=1+n\alpha+\frac{n(n-1)}2\,\alpha^2. $$

Answer 2

There's almost nothing to prove. $(1+\alpha)^n=\binom{n}{0}+\binom{n}{1}\alpha+\binom{n}{2}\alpha^2+\binom{n}{3}\alpha^3+\cdots +\alpha^n = 1 + n\alpha +\frac{n(n-1)}{2}\alpha^2 + \binom{n}{3}\alpha^3+\cdots +\alpha^n >= 1 + n\alpha +\frac{n(n-1)}{2}\alpha^2 $

If n = 1 or 2 it's an equality. If n > 2 it's an inequality.