Find N? Please add work and explanation!

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Find N? Please add work and explanation!

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

1 Answer

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Answer 1 · 2017-05-12T22:29:21

$1000 \sum k = 1 k (⌈ {log}_{\sqrt{2}} k ⌉ - ⌊ {log}_{\sqrt{2}} k ⌋) = 499477$ $sum_(k=1)^1000 k(ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k))=499477$

Explanation:

When is ${log}_{\sqrt{2}} k$ $log_(sqrt(2)) k$ an integer?

If $k$ $k$ is a power of $2$ $2$ , e.g. $k = 2^{m}$ $k = 2^m$ then we find:

${log}_{\sqrt{2}} k = {log}_{\sqrt{2}} 2^{m} = 2 m$ $log_(sqrt(2)) k = log_(sqrt(2)) 2^m = 2m$

Conversely, if ${log}_{\sqrt{2}} k$ $log_(sqrt(2)) k$ is an integer, e.g. $n$ $n$ , then we find:

$k = {\sqrt{2}}^{n} = 2^{\frac{n}{2}}$ $k = sqrt(2)^n = 2^(n/2)$

If $n$ $n$ is odd then this is irrational and not an integer.

Note that when ${log}_{\sqrt{2}} k$ $log_(sqrt(2)) k$ is an integer, then:

$⌈ {log}_{\sqrt{2}} k ⌉ - ⌊ {log}_{\sqrt{2}} k ⌋ = 0$ $ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k) = 0$

When ${log}_{\sqrt{2}} k$ $log_(sqrt(2)) k$ is not an integer, then:

$⌈ {log}_{\sqrt{2}} k ⌉ - ⌊ {log}_{\sqrt{2}} k ⌋ = 1$ $ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k) = 1$

So we find:

$1000 \sum k = 1 k (⌈ {log}_{\sqrt{2}} k ⌉ - ⌊ {log}_{\sqrt{2}} k ⌋)$ $sum_(k=1)^1000 k(ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k))$

$= 1000 \sum k = 1 k - 9 \sum m = 0 2^{m}$ $=sum_(k=1)^1000 k - sum_(m=0)^9 2^m$

$= \frac{1}{2} \cdot 1000 \cdot (1000 + 1) - (2^{10} - 1)$ $=1/2*1000*(1000+1) - (2^10-1)$

$= 500500 - 1023$ $=500500-1023$

$= 499477$ $=499477$

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Find N? Please add work and explanation!

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

1 Answer

Explanation:

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Find N? Please add work and explanation!

Let N = \sum_{k = 1}^{1000}k(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor). N=1000∑k=1k(⌈log√2k⌉−⌊log√2k⌋).N = \sum_{k = 1}^{1000}k(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor). Find N

1 Answer

Explanation:

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