# Math Help - one step to the other..how?

1. ## one step to the other..how?

im reviewing a proof by induction and wondering how

$2^{n+1} >2n^{2} = (n+1)^{2} + (n-2n-1)$

the original question was prove $2^{n} >n^{2}$ for all n $\geq 5$

thanks

2. Hello,

Originally Posted by skystar

$2n^{2} = (n+1)^{2} + (n-2n-1)$
$(n+1)^2=n^2+2n+1$

Hence $(n+1)^{2} + (n-2n-1)=2n^2$

$2^{n+1} >2n^{2}$
By multiplying the hypothesis : $2^n>n^2$ by 2