The solutions manual gives this equality without an explanation (as usual); I've been stuck at this for a good while and don't feel like wasting any more time $$\frac{(n+1)^2}{2^{n+1}}\times\frac{2^n}{n^2}=\left(1+\frac{1}{n}\right)^2\times\frac{1}{2}$$
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4Hint: $;2^{n+1} = 2 \cdot 2^n,$ and $,\frac{n+1}{n} = \frac{n}{n} + \frac{1}{n},$. – dxiv May 02 '23 at 03:47
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1Why is this tagged with [inequality]? – Martin R May 02 '23 at 05:18