what the following may equal? ∑ i = 1 n i ⌈ log 2 ⁡ ( i ) ⌉ = ?

Question

what the following may equal?      ∑          i      =      1              n        i      ⌈          log      2        ⁡          (      i      )        ⌉    =  ?

faux0101d · Accepted Answer

Use sum by parts:      ∑          1      ≤      k      ≤      n            x    k    Δ      y    k    =      x          n      +      1            y          n      +      1        −      x    1        y    1    −      ∑          1      ≤      k      ≤      n            y          k      +      1        Δ      x    k  In your case   Δ      y    k    =  ⌈      log    2    ⁡  k  ⌉ and       x    k    =  k. You&#039;ll get a sum similar to the original, you should be able to solve for that one.

caschaillo7 · Accepted Answer

I could come with a solution with an incisive precision for any base a by doing this:

T (n) = \sum_{i = 1}^{n} i ⌈ \log_{a} (i) ⌉ = [\sum_{i = 1}^{⌊ \log_{a} (n) ⌋} i (\sum_{j = a^{i - 1} + 1}^{a^{i}} j)] + ⌈ \log_{a} (n) ⌉ \sum_{i = a^{⌊ \log_{a} (n) ⌋} + 1}^{n} i T (n) = [\frac{1}{2} \sum_{i = 1}^{⌊ \log_{a} (n) ⌋} i (a^{i - 2} (a - 1) (a^{i + 1} + a^{i} + a))] + ⌈ \log_{a} (n) ⌉ \sum_{i = a^{⌊ \log_{a} (n) ⌋} + 1}^{n} i Let T_{1} (n) = [\frac{1}{2} \sum_{i = 1}^{⌊ \log_{a} (n) ⌋} i (a^{i - 2} (a - 1) (a^{i + 1} + a^{i} + a))] And T_{2} (n) = ⌈ \log_{a} (n) ⌉ \sum_{i = a^{⌊ \log_{a} (n) ⌋} + 1}^{n} i T_{1} (n) = \frac{1}{2 (a^{2} - 1)} ((- 1 - ⌊ \log_{a} (n) ⌋) a^{2 ⌊ \log_{a} (n) ⌋} - a^{⌊ \log_{a} (n) ⌋ + 1} + ⌊ \log_{a} (n) ⌋ a^{⌊ \log_{a} (n) ⌋ + 2} + ⌊ \log_{a} (n) ⌋ a^{2 ⌊ \log_{a} (n) ⌋ + 2} - (⌊ \log_{a} (n) ⌋ + 1) a^{⌊ \log_{a} (n) ⌋} + a + 2) T_{2} (n) = \frac{⌈ \log_{a} (n) ⌉ (n - a^{⌊ \log_{a} (n) ⌋}) (a^{⌊ \log_{a} (n) ⌋} + n + 1)}{2} Therefore T (n) = T_{1} (n) + T_{2} (n)

what the following may equal? sum_(i=1)^(n) [log_2(i)]=?

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New Questions in Calculus 1