\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
Tính A = .........
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MT
29 tháng 12 2015
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+99}+\frac{1}{50}\)
\(=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}=2.\left(\frac{50}{100}-\frac{1}{100}\right)+\frac{1}{50}=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=1\)
PT
2
31 tháng 12 2015
Ta có: A=(1/1+2)+(1/1+2+3)+...+(1/1+2+3+...+99)+(1/50)
A=[1/(2+1).2/2]+[1/(1+3).3/2]+....+[1/(1+99).99/2]+(1/50)
A= [2/(2+1).2]+[2/(1+3).3)]+...+[2/(1+99).99]+1/50)
A=2.[(1/2.3)+(1/3.4)+...+(1/99.100)]+(1/50)
A=2.(1/2-1/3+1/4-1/4+...+1/99-1/100)+(1/50)
A=2.(1/2-1/100)+(1/50)
A=2.(49/100)+(1/50)
A=1
đảm bảo đọc k hiểu
A = \(\dfrac{1}{1+2}\) + \(\dfrac{1}{1+2+3}\) + ... + \(\dfrac{1}{1+2+3+...+99}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{1}{\left(2+1\right).2:2}\) + \(\dfrac{1}{\left(3+1\right).3:2}\) + ... + \(\dfrac{1}{\left(99+1\right).99:2}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{99.100}\) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}-\dfrac{1}{5}\)+ \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.(\(\dfrac{50}{100}\) - \(\dfrac{1}{100}\)) + \(\dfrac{1}{50}\)
A = 2.\(\dfrac{49}{100}\) + \(\dfrac{1}{50}\)
A = \(\dfrac{49}{50}\) + \(\dfrac{1}{50}\)
A = 1