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

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.

• 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 $0$. – André Nicolas Feb 23 '15 at 1:02

## 3 Answers

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 $\left\lceil\log_{\sqrt2}k\right\rceil-\left\lfloor\log_{\sqrt2}k\right\rfloor$

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 $1$

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:

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

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