\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)\)
= \(2\times(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51})\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{51}\right)\)
= \(2\times\frac{49}{102}\)
= \(\frac{49}{51}\)
A=1/1+2 + 1/1+2+3 + 1/1+2+3+4 +... + 1/1+2+3+...+50
A = 1/3 + 1/6 + 1/10 + 1/15 + ...+1/1275
Nhân cả hai vế với 1/2, ta có:
A/2 = 1/6 + 1/12 + 1/20 + 1/30 + ... + 1/2550
A/2 = 1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + ... + 1/50x51
A/2 = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +..... + 1/50 - 1/51
A/2 = 1-1/51
A/2 = 49/102
A = 49/51
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
\(=\frac{2}{2(1+2)}+\frac{2}{2(1+2+3)}+\frac{2}{2(1+2+3+4)}+...+\frac{2}{2(1+2+3+...+50)}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2\left[\frac{(1+50)\cdot50}{2}\right]}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2550}\)
\(=2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right]\)
\(=2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{50\cdot51}\right]\)
\(=2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right]\)
\(=2\left[\frac{1}{2}-\frac{1}{51}\right]=\frac{49}{51}\)
\(\frac{1}{1+2}\)+\(\frac{1}{1+2+3}\)+\(\frac{1}{1+2+3+4}\)+....+\(\frac{1}{1+2+3+4+....+50}\)
B=\(\frac{1}{\left[1+2\right]\cdot2:2}\)+\(\frac{1}{\left[1+3\right]\cdot3:2}\)+\(\frac{1}{\left[1+4\right]\cdot4:2}\)+....+\(\frac{1}{\left[1+50\right]\cdot50:2}\)
B=\(\frac{1}{3\cdot2:2}\)+\(\frac{1}{4\cdot3:2}\)+\(\frac{1}{5\cdot4:2}\)+....+\(\frac{1}{51\cdot50:2}\)
B*\(\frac{1}{2}\)=\(\frac{1}{3\cdot2}\)+\(\frac{1}{4\cdot3}\)+\(\frac{1}{5\cdot4}\)+....+\(\frac{1}{51\cdot50}\)
B*\(\frac{1}{2}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{50}\)-\(\frac{1}{51}\)
B*\(\frac{1}{2}\)=\(\frac{1}{2}\)-\(\frac{1}{51}\)
B*\(\frac{1}{2}\)=\(\frac{49}{102}\)
B=\(\frac{49}{102}\):\(\frac{1}{2}\)
B=\(\frac{49}{51}\)
Dấu . là Dấu nhân
Đặt \(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+50}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{50\times51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{50}-\frac{1}{50}\right)-\frac{1}{51}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{51}=\frac{49}{102}\)
\(\Rightarrow A=\frac{49}{102}:\frac{1}{2}=\frac{49}{51}\)
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