\(\frac{1}{3}\)+\(\frac{1}{6}\).(1+2)+\(\frac{1}{9}\).(1+2+3)+.....+ \(\frac{1}{6045}\).(1+2+3+...+2015)= B
tính= cách hợp lý
Những câu hỏi liên quan
Tính B=$\frac{1}{3} \frac{1}{6}\left(1 2\right) \frac{1}{9}\left(1 2 3\right) ... \frac{1}{6045}\left(1 2 3 ... 2015\right)$13 16 (1 2) 19 (1 2 3) ... 16045 (1 2 3 ... 2015)
Thực hiện phép tính bằng cách hợp lí
A=\(\frac{\frac{3}{11}+1-\frac{3}{7}}{3+\frac{9}{11}-\frac{9}{7}}-\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}\)
B = \(\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+...+\frac{1}{6045}.\left(1+2+3+...+2015\right)\)
BÀI 1:TÍNH:
\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+....+\frac{100}{2^{100}}\)
BÀI 2: CHỨNG MINH RẰNG:
\(B=1-\frac{1}{2^2}-\frac{1}{3^2}-.....-\frac{1}{2004^2}>\frac{1}{2004}\)
BÀI 3:THỰC HIỆN PHÉP TÍNH BẰNG CÁCH HỢP LÝ:
\(B=\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+.....+\frac{1}{6045}.\left(1+2+3+....+2015\right)\)
Tính A=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
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Tính B=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+3+...+2015\right)\)
Ta có: \(1+2+...+n=\frac{n\left(n+1\right)}{2}\) áp dụng vào bài toán ta được
\(B=\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+...+2015\right)\)
\(=\frac{1}{3}+\frac{1}{2.3}.\frac{2.3}{2}+\frac{1}{3.3}.\frac{3.4}{2}+...+\frac{1}{2015.3}.\frac{2015.2016}{2}\)
\(=\frac{1}{3}\left(1+\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\right)\)
\(=\frac{1}{6}\left(2+3+4+...+2016\right)=\frac{1}{6}.\frac{2015.2018}{2}=\frac{2033135}{6}\)
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B=\(\frac{1}{3}+\frac{1}{6}\cdot\left(1+2\right)+\frac{1}{9}\cdot\left(1+2+3\right)+...+\frac{1}{6045}\cdot\left(1+2+3+...+2015\right)\)
Tính \(B=\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}.\left(1+2+3\right)+...+\frac{1}{6045}.\left(1+2+..+2015\right)\)
Tính B=\(\frac{1}{3}+\frac{1}{6}\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+...+\frac{1}{6045}\left(1+2+3+...+2015\right)\)
Tính B=\(\frac{1}{3}+\frac{1}{6}.\left(1+2\right)+\frac{1}{9}\left(1+2+3\right)+....+\frac{1}{6045}\left(1+2+3+\text{4}+.....+2015\right)\)
CMCT : ( tự CM )
Áp dụng bài toán trên ta có : \(B=\frac{1}{3}+\frac{1}{6}\vec{\left(\frac{\left(x+1\right).2}{2}\right)+\frac{1}{9}\left(\frac{\left(1+3\right).3}{2}\right)+...+}\frac{1}{6045}\left(\frac{\left(1+2015\right).2}{2}\right)\)
\(B=\frac{1}{3}+\frac{1}{2}+\frac{2}{3}+....+....\) ( tự lm tiếp )
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