Identitatea lui Hermite:
[tex][x]+\left[x+\frac{1}{2}\right] = [2x],\,\,\forall x\in \mathbb{R}[/tex]
Rezolvare:
[tex]S = \left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\left[\frac{n+4}{8}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right][/tex]
[tex]\Rightarrow S = \left[\frac{n}{2}+\frac{1}{2}\right]+\left[\frac{n}{4}+\frac{2}{4}\right]+\left[\frac{n}{8}+\frac{4}{8}\right]+...+\left[\frac{n}{2^{k+1}}+\frac{2^k}{2^{k+1}}\right][/tex]
[tex]\Rightarrow S = \left[\frac{n}{2}+\frac{1}{2}\right]+\left[\frac{n}{4}+\frac{1}{2}\right]+\left[\frac{n}{8}+\frac{1}{2}\right]+...+\left[\frac{n}{2^{k+1}}+\frac{1}{2}\right][/tex]
[tex]\Rightarrow S = \left(\left[\frac{n}{2}\right]+\left[\frac{n}{2}+\frac{1}{2}\right]-\left[\frac{n}{2}\right]\right)+\left(\left[\frac{n}{4}\right]+\left[\frac{n}{4}+\frac{1}{2}\right]-\left[\frac{n}{4}\right]\right)+...\\+\left(\left[\frac{n}{2^{k+1}}\right]+\left[\frac{n}{2^{k+1}}+\frac{1}{2}\right]-\left[\frac{n}{2^{k+1}}\right]\right)[/tex]
[tex]\Rightarrow S = \left(\left[2\cdot \frac{n}{2}\right]-\left[\frac{n}{2}\right]\right)+\left(\left[2\cdot \frac{n}{4}\right]-\left[\frac{n}{4}\right]\right)+\left(\left[2\cdot \frac{n}{8}\right]-\left[\frac{n}{8}\right]\right)+...\\+\left(\left[2\cdot \frac{n}{2^{k}}\right]-\left[\frac{n}{2^{k}}\right]\right)+\left(\left[2\cdot \frac{n}{2^{k+1}}\right]-\left[\frac{n}{2^{k+1}}\right]\right)[/tex]
[tex]\Rightarrow S = [n] - \left[\frac{n}{2}\right] + \left[\frac{n}{2}\right]- \left[\frac{n}{4}\right] + \left[\frac{n}{4}\right] - \left[\frac{n}{8}\right]+...+\left[\frac{n}{2^{k-1}}\right]-\left[\frac{n}{2^k}\right]+\\+\left[\frac{n}{2^k}\right] - \left[\frac{n}{2^{k+1}}\right][/tex]
[tex]\Rightarrow S = [n]-\left[\frac{n}{2^{k+1}}\right],\,\,\forall n, k \in \mathbb{N},\,\, k<n[/tex]
#copaceibrainly
