Ta có: với \(x^2\le n< \left(x+1\right)^2\Rightarrow\left[n\right]=x\)
Do đó ta chia tổng A thành các nhóm:
\(\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]=3.1\)
\(\left[\sqrt{4}\right]+\left[\sqrt{5}\right]+...+\left[\sqrt{8}\right]=5.2\)
\(\left[\sqrt{9}\right]+\left[\sqrt{10}\right]+...+\left[\sqrt{15}\right]=7.3\)
...
\(\left[\sqrt{169}\right]+\left[\sqrt{170}\right]+...+\left[\sqrt{195}\right]=27.13\)
\(\left[\sqrt{196}\right]+...+\left[\sqrt{200}\right]=5.14\)
Do đó:
\(A=3.1+5.2+7.3+...+27.13+5.14=1799\)
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