JEEOlympiad
Compute sum_{n=1}^{1023} floor(log_2 n), where floor is the greatest…
Question
Compute sum_{n=1}^{1023} floor(log_2 n), where floor is the greatest integer function and log_2 is the base-2 logarithm.
✓ Verified answer: 8194checked by our engine — not a guess
Step-by-step solution
For 2^k <= n <= 2^{k+1} - 1 we have floor(log_2 n) = k, and there are 2^k such integers.
For n from 1 to 1023 = 2^10 - 1, k runs from 0 to 9.
Sum = sum_{k=0}^{9} k*2^k. Using sum_{k=0}^{m} k*2^k = (m-1)2^{m+1} + 2 with m = 9:
= 8*2^10 + 2 = 8*1024 + 2 = 8192 + 2 = 8194.
Final answer8194
Stuck on a problem like this?
Paste any JEE or NEET question — verified working, a confidence %, and an honest “not sure” instead of a bluff.
Solve my doubt →More Sequences Series solutions
Compute floor(1000 * T), where T = sum_{n=0}^{infinity} (n^2 + 1)/n!…JEE · MathThe sum S = sum_{n=1}^{80} 1/((n+1)*sqrt(n) + n*sqrt(n+1)) equals m/n…JEE · MathThe infinite sum sum_{n=1}^{infinity} n/(n^4 + n^2 + 1) equals m/n in…JEE · MathLet F_1 = 1, F_2 = 1 and F_{n+2} = F_{n+1} + F_n be the Fibonacci…JEE · MathThe infinite sum sum_{n=1}^{infinity} 1/(n(n+2)(n+4)) equals m/n in…JEE · Math