Tính 1+1/2+(1+2)+1/3(1+2+3)+...+1/2016(1+2+3+...+2016)
Giúp mình làm bài này nhé! 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Có kết quả và các bước cụ thể giùm mình nha! Cảm ơn các bạn nhiều!
Trả lời (1)
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Ta có công thức:
\(1+2+3+...+n=\dfrac{n\cdot\left(n+1\right)}{2}\)
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{2016}\left(1+2+3+...+2016\right)\\ =1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{2016}\cdot\dfrac{2016\cdot2017}{2}\\ =1+\dfrac{1\cdot2\cdot3}{2\cdot2}+\dfrac{1\cdot3\cdot4}{3\cdot2}+...+\dfrac{1\cdot2016\cdot2017}{2016\cdot2}\\ =\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{2017}{2}\\ =\dfrac{2+3+4+...+2017}{2}\\ =\dfrac{1+2+3+...+2017-1}{2}\\ =\dfrac{\dfrac{2017\cdot2018}{2}-1}{2}\\ =\dfrac{2035153-1}{2}\\ =\dfrac{2035152}{2}\\ =1017576\)
bởi Đậu Lâm
19/12/2018
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