Question

Salut! Putin ajutor, clasa a 9a, exercitii cu partea intreaga. Multumesc anticipat! (paranteza patrata->parte intreaga)


Calculati suma:

$S=[\frac{n+1}{2} ]+[\frac{n+2}{4} ]+[\frac{n+4}{8} ]+...+[\frac{n+2^K}{2^{K+1}} ]$
n∈|N; K

Answer 1

Identitatea lui Hermite:

$[x]+\left[x+\frac{1}{2}\right] = [2x],\,\,\forall x\in \mathbb{R}$

Rezolvare:

$S = \left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\left[\frac{n+4}{8}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right]$

$\Rightarrow S = \left[\frac{n}{2}+\frac{1}{2}\right]+\left[\frac{n}{4}+\frac{2}{4}\right]+\left[\frac{n}{8}+\frac{4}{8}\right]+...+\left[\frac{n}{2^{k+1}}+\frac{2^k}{2^{k+1}}\right]$

$\Rightarrow S = \left[\frac{n}{2}+\frac{1}{2}\right]+\left[\frac{n}{4}+\frac{1}{2}\right]+\left[\frac{n}{8}+\frac{1}{2}\right]+...+\left[\frac{n}{2^{k+1}}+\frac{1}{2}\right]$

$\Rightarrow S = \left(\left[\frac{n}{2}\right]+\left[\frac{n}{2}+\frac{1}{2}\right]-\left[\frac{n}{2}\right]\right)+\left(\left[\frac{n}{4}\right]+\left[\frac{n}{4}+\frac{1}{2}\right]-\left[\frac{n}{4}\right]\right)+...\\+\left(\left[\frac{n}{2^{k+1}}\right]+\left[\frac{n}{2^{k+1}}+\frac{1}{2}\right]-\left[\frac{n}{2^{k+1}}\right]\right)$

$\Rightarrow S = \left(\left[2\cdot \frac{n}{2}\right]-\left[\frac{n}{2}\right]\right)+\left(\left[2\cdot \frac{n}{4}\right]-\left[\frac{n}{4}\right]\right)+\left(\left[2\cdot \frac{n}{8}\right]-\left[\frac{n}{8}\right]\right)+...\\+\left(\left[2\cdot \frac{n}{2^{k}}\right]-\left[\frac{n}{2^{k}}\right]\right)+\left(\left[2\cdot \frac{n}{2^{k+1}}\right]-\left[\frac{n}{2^{k+1}}\right]\right)$

$\Rightarrow S = [n] - \left[\frac{n}{2}\right] + \left[\frac{n}{2}\right]- \left[\frac{n}{4}\right] + \left[\frac{n}{4}\right] - \left[\frac{n}{8}\right]+...+\left[\frac{n}{2^{k-1}}\right]-\left[\frac{n}{2^k}\right]+\\+\left[\frac{n}{2^k}\right] - \left[\frac{n}{2^{k+1}}\right]$

$\Rightarrow S = [n]-\left[\frac{n}{2^{k+1}}\right],\,\,\forall n, k \in \mathbb{N},\,\, k<n$

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