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
George C.
May 12, 2017

#sum_(k=1)^1000 k(ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k))=499477#

Explanation:

When is #log_(sqrt(2)) k# an integer?

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

#log_(sqrt(2)) k = log_(sqrt(2)) 2^m = 2m#

Conversely, if #log_(sqrt(2)) k# is an integer, e.g. #n#, then we find:

#k = sqrt(2)^n = 2^(n/2)#

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

Note that when #log_(sqrt(2)) k# is an integer, then:

#ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k) = 0#

When #log_(sqrt(2)) k# is not an integer, then:

#ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k) = 1#

So we find:

#sum_(k=1)^1000 k(ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k))#

#=sum_(k=1)^1000 k - sum_(m=0)^9 2^m#

#=1/2*1000*(1000+1) - (2^10-1)#

#=500500-1023#

#=499477#

Related questions
Impact of this question
1064 views around the world
You can reuse this answer
Creative Commons License