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How do you prove in induction that: $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$

This is what I keep getting:

Checking for $n=1$ we get $2\le\frac{9}{4}$.

Assuming it's true for $n$ and checking for $n+1$ I get this: $\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\times2^n}\le\frac{9}{4}-\frac{n^2-2}{2\times2^n}\le\frac{9}{4}$ Which is true only for $n>1$.

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    @Henry, I see, you are right. In this case, the induction basis must be done for$n=1$and n=2, and only then you should do the induction step.2012-11-13

2 Answers 2

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The simple answer is to check $n=2$ too, where you have

$\frac{(n+1)^2}{2^n}= \frac{(2+1)^2}{2^2} = \frac{9}{4} \le\frac{9}{4}$ and then do the induction.

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For another proof, you can see that $\displaystyle \frac{u_{n+1}}{u_n}= \frac{1}{2} \left( 1+ \frac{1}{n+1} \right)^2 \leq 1$ with $n\geq 2$ and $\displaystyle u_n= \frac{(n+1)^2}{2^n}$. So $(u_n)$ decreases for $n\geq 2$ whence $\displaystyle u_n \leq u_2=\frac{9}{4}$ for $n \geq 2$. Finally, you only have to check the case $n=1$.