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Find $N$ for

$$N =\sum_{k = 1}^{1000}k\left(\left\lceil\log_{\sqrt{2}}k\right\rceil-\left\lfloor\log_{\sqrt{2}}k\right\rfloor\right)\;.$$

How could you solve this problem? Are there sigma rules or anything? Thanls.

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    $\begingroup$ The term after the $k$ is usually $1$, and occasionally $0$. It seems sensible to pretend they are all $1$. That gives a familiar sum. Then compensate for the cases they are $0$. $\endgroup$ – André Nicolas Feb 23 '15 at 1:02
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HINT: If $k=2^n$, then $\log_{\sqrt2}k=2n$, and therefore

$$\left\lceil\log_{\sqrt2}k\right\rceil=\left\lfloor\log_{\sqrt2}k\right\rfloor\;.\tag{1}$$

  • Are there any other values of $k$ for which $(1)$ is true?
  • When $(1)$ is false, what is $\left\lceil\log_{\sqrt2}k\right\rceil-\left\lfloor\log_{\sqrt2}k\right\rfloor$
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Hint:
When $k$ is a power of $2$ then $\left\lceil\log_{\sqrt{2}}k\right\rceil-\left\lfloor\log_{\sqrt{2}}k\right\rfloor = 0$ otherwise it's $1$

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Just see this if $\log_{\sqrt{2}}k$ is not integer then

$$ \left\lceil\log_{\sqrt{2}}k\right\rceil-\left\lfloor\log_{\sqrt{2}}k\right\rceil = 1$$

and $$ \left\lceil\log_{\sqrt{2}}k\right\rceil-\left\lfloor\log_{\sqrt{2}}k\right\rceil = 0$$

otherwise.

Note:

$$ \lceil x \rceil - \lfloor x \rfloor = \begin{cases}0 ,\quad z\in \mathbb{Z}\\ 1\quad otherwise \end{cases} $$

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  • $\begingroup$ But are the logs never integers? $\endgroup$ – Ross Millikan Feb 23 '15 at 1:11

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