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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.

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The key is that $$ {n\choose2}2^2=\frac{4n!}{(n-2)!2!}=2n(n-1). $$ Then you have, since all terms are positive, $$ (1+2)^n=1+2n+2n(n-1)+\cdots=1+2n^2+\cdots\geq1+2n^2. $$

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$$\begin{align} (2+1)^n & = \color{red}{1}+2n+2\color{red}n(\color{red}n-1)+\text{stuff} \\ & = 1+2n^2+2n-2n+\text{stuff} \\ & = 1+2n^2+\text{stuff} \\ \end{align}$$

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.

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