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
Let $$N = \sum_{k = 1}^{1000}k(\lceil \log_{\sqrt {2}}k\rceil - \lfloor \log_{\sqrt {2}}k \rfloor). $$
Find N
1 Answer
#sum_(k=1)^1000 k(ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k))=499477#
Explanation:
When is
If
#log_(sqrt(2)) k = log_(sqrt(2)) 2^m = 2m#
Conversely, if
#k = sqrt(2)^n = 2^(n/2)#
If
Note that when
#ceil(log_(sqrt(2)) k) - floor(log_(sqrt(2)) k) = 0#
When
#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#