# Prove the inequality using the binomial theorem

Prove $3^n \geq 2n^2 +1$ for $n = 1,2,\ldots$ using binomial theorem, applied to $(1+x)^n$ with $x=2$.

So I started with expanding $(1+2)^n$ = ${n \choose 0} + {n \choose 1}2 + {n \choose 2}2^2 + \cdots + {n \choose n} \geq 1 + 2n! + 2^n$

I'm not sure what I'm doing wrong or what the next step should be - any help/hints will be appreciated.

The key is that $${n\choose2}2^2=\frac{4n!}{(n-2)!2!}=2n(n-1).$$ Then you have, since all terms are positive, $$(1+2)^n=1+2n+2n(n-1)+\cdots=1+2n^2+\cdots\geq1+2n^2.$$
\begin{align} (2+1)^n & = \color{red}{1}+2n+2\color{red}n(\color{red}n-1)+\text{stuff} \\ & = 1+2n^2+2n-2n+\text{stuff} \\ & = 1+2n^2+\text{stuff} \\ \end{align}
Note that since we are binomial expanding for positive numbers, stuff is positive and greater than $0$, assuming the binomial expansion goes out that many terms.