Giúp mình với
1/2+1/3+1/4+...+1/100
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A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
2A = \(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{99}}\)
A = 2A - A = \(1-\frac{1}{2^{100}}
\(\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+...+\dfrac{1}{y.(y+3)}=\dfrac{98}{1545}\)
\(3.(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{y(y+3)})=\dfrac{98}{1545}.3\)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{y(y+3)}=\dfrac{98}{515}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(y+3.1=1.515\)
\(y+3=515\)
\(y=515-3\)
\(y=512\)
Vậy y = 512
Nhớ tick cho mk nha
ta có A = 1+(1+2)+....+(1+2+..+100) = 1 x 100 + 2 x 99 + ...+100 x 1
\(\Rightarrow\frac{A}{100.1+99.2+...+1.100}=\frac{100.1+99.2+..+1.100}{100.1+99.2+..+100.1}=1\)
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\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\) mà