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b1
a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{5}-\dfrac{1}{10}\)
\(=\dfrac{2}{10}-\dfrac{1}{10}\)
\(=\dfrac{1}{10}\)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{1}-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)
\(=\dfrac{1}{3}-\dfrac{1}{11}\)
\(=\dfrac{8}{33}\)
d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{98}{303}\)
\(a,A=2^1+2^2+2^3+...+2^{2019}\)
\(2A=2^2+2^3+2^4+...+2^{2020}\)
\(\Rightarrow2A-A=A=2^{2020}-2\)
\(B=1+3+3^2+3^3+...+3^{2020}\)
\(3B=3+3^2+3^3+...+3^{2021}\)
\(3B-B=2B=3^{2021}-1\)
\(B=\frac{3^{2021}-1}{2}\)
a,\(A=2^1+2^2+2^3+...+2^{2019}\)
\(2A=2^2+2^3+2^4+...+2^{2020}\)
\(2A-A=\left[2^2+2^3+2^4+...+2^{2020}\right]-\left[2^1+2^2+...+2^{2019}\right]\)
\(A=2^{2020}-2^1=2^{2020}-2\)
b, \(B=1+3+3^2+3^3+...+3^{2020}\)
\(3B=3+3^2+3^3+...+3^{2021}\)
\(3B-B=\left[3+3^2+3^3+...+3^{2021}\right]-\left[1+3+3^2+...+3^{2020}\right]\)
\(2B=3^{2021}-1\)
\(B=\frac{3^{2021}-1}{2}\)
1.a
1/2+1/4+1/8+1/16+1/32
= 1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
= 1-1/32=31/32
1b
\(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3} +\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{2}{3}+\frac{1}{4}+\frac{1}{20}+\frac{1}{30}\)
\(=\frac{5}{20}+\frac{5}{30}+\frac{20}{30}+\frac{5}{20}+\frac{1}{20}+\frac{1}{30}\)
\(=\left(\frac{5}{20}+\frac{5}{20}+\frac{1}{20}\right)+\left(\frac{5}{30}+\frac{20}{30}+\frac{1}{30}\right)\)
\(=\frac{11}{20}+\frac{26}{30}\)
\(=\frac{11}{20}+\frac{13}{15}\)
\(=\frac{17}{12}\)