im reviewing a proof by induction and wondering how $\displaystyle 2^{n+1} >2n^{2} = (n+1)^{2} + (n-2n-1) $ the original question was prove $\displaystyle 2^{n} >n^{2}$ for all n $\displaystyle \geq 5$ thanks
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Hello, Originally Posted by skystar $\displaystyle 2n^{2} = (n+1)^{2} + (n-2n-1) $ $\displaystyle (n+1)^2=n^2+2n+1$ Hence $\displaystyle (n+1)^{2} + (n-2n-1)=2n^2$ $\displaystyle 2^{n+1} >2n^{2}$ By multiplying the hypothesis : $\displaystyle 2^n>n^2$ by 2 Does it help you understand ?
but would that go to n^2 +n confused still
Originally Posted by skystar but would that go to n^2 +n confused still Where ? oO
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