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Với mọi \(x\in Z\) ta có:
\(1+2+3+..+n=\frac{n\left(n+1\right)}{2}\)
=> \(\frac{1}{1+2+3+..+n}=\frac{2}{n\left(n+1\right)}=2\left[\frac{1}{n\left(n+1\right)}\right]=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)
Có:
\(\frac{1}{1+2}=2\left(\frac{1}{2}-\frac{1}{3}\right)\)
\(\frac{1}{1+2+3}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)
.......................................................
\(\frac{1}{1+2+3+4+...+99}=2\left(\frac{1}{99}-\frac{1}{100}\right)\)
Nên:
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+..+99}+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+...+2\left(\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2\cdot\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=1\)
Cảm ơn bạn (chị ) nhiều !
Công nhận chị học giỏi thật đấy !
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\left(\frac{50}{100}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=1\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+\frac{1}{\frac{\left(4+1\right).4}{2}}+....+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{1}{\frac{3.2}{2}}+\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+....+\frac{1}{\frac{100.99}{2}}+\frac{1}{50}\)
\(=\frac{2}{3.2}+\frac{2}{4.3}+\frac{2}{5.4}+...+\frac{2}{100.99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}=\frac{49}{50}+\frac{1}{50}=1\)
\(A=\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{99}-\frac{2}{100}+\frac{1}{50}\)
\(A=\frac{2}{2}-\frac{2}{100}+\frac{1}{50}=1\)