Compute a summation with limit log

Question

I want to compute the summation of

$$ \sum_{i=1}^{\lfloor{\log(n)}\rfloor{}}{i} $$

I could not find an axiom to solve it. thank!

Answer 1

See that

In other words,

$$2(1+2+3+\dots+n)=n(n+1)$$

$$1+2+3+\dots+n={n(n+1)\over2}$$

So, plugging our stuff in, we have

$$1+2+3+\dots+\lfloor\log(n)\rfloor={\lfloor\log(n)\rfloor\lfloor1+\log(n)\rfloor\over2}$$