k-1-n-log-2-k-2018-What-is-the-value-of-n-

Question Number 9516 by Joel575 last updated on 12/Dec/16

\underset{k = 1}{\sum^{n}} ⌊ \log_{2} k ⌋ = 2018

What is the value of n ?

Commented by sou1618 last updated on 12/Dec/16

* K = ⌊ {l o g}_{2} k ⌋

w h e n 2^{0} ⩽ k < 2^{1} k = {1}

K = 0

w h e n 2^{1} ⩽ k < 2^{2} k = {2, 3}

K = 1

\dots

(m = 0, 1, 2, 3 \dots .)

w h e n 2^{m} ⩽ k < 2^{m + 1} k = {2^{m}, 2^{m} + 1, 2^{m} + 2, \dots, 2^{m + 1} - 1}

K = m

T_{m} = \underset{k = 2^{m}}{\sum^{(2^{m + 1} - 1)}} K

= \underset{k = 2^{m}}{\sum^{(2^{m + 1} - 1)}} m = m 2^{m}

- - - - - - - - - -

\underset{k = 1}{\sum^{n}} K = \underset{m = 1}{\sum^{⌊ {l o g}_{2} n ⌋}} (\underset{k = 2^{m - 1}}{\sum^{2^{m} - 1}} K) + \underset{(k = 2^{⌊ {l o g}_{2} n ⌋})}{\sum^{n}} K

\underset{k = 1}{\sum^{n}} K = \underset{m = 0}{\sum^{⌊ {l o g}_{2} n ⌋ - 1}} T_{m} + \underset{(k = 2^{⌊ {l o g}_{2} n ⌋})}{\sum^{n}} K

s o f i n d N : s a t i s f y i n g ↓

\underset{m = 0}{\sum^{N}} T_{m} < 2018

= 0 \times 2^{0} + 1 \times 2^{1} + 2 \times 2^{2} + 3 \times 2^{3} \dots + N \times 2^{N} < 2018

0 + 2 + 8 + 24 + 64 + 160 + 384 + 896 = 1538 < 2018

∴ N = 7

\Rightarrow 7 = ⌊ {l o g}_{2} n ⌋ - 1 \Leftrightarrow 2^{8} ⩽ n < 2^{9}

\underset{k = 2^{8}}{\sum^{n}} K = 2018 - 1538 = 480

w h e n 2^{8} ⩽ n < 2^{8 + 1}

K = 8

\Rightarrow

\underset{k = 2^{8}}{\sum^{n}} K = (n + 1 - 2^{8}) \times 8 = 480

∴ n = 2^{8} + 59

Commented by Joel575 last updated on 14/Dec/16

t h a n k y o h v e r y m u c h

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