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\(B=10101\cdot\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\left(\dfrac{2}{222222}+\dfrac{5}{222222}\right)\)
\(=10101\cdot\dfrac{1}{31746}=\dfrac{7}{22}\)
10101 . \(\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= 10101 . \(\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= 10101 . \(\frac{7}{222222}\)
= \(\frac{7}{22}\)
Lời giải:
Ta có:
\(A=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
\(=10101.\left(\frac{10}{222222}+\frac{5}{222222}-\frac{8}{222222}\right)\)
\(=10101.\left(\frac{10+5-8}{222222}\right)\)
\(=10101.\frac{7}{222222}\)
\(=\frac{7.10101}{22.10101}=\frac{7}{22}\)
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\(C=\frac{1.2.3+2.3.4+3.4.5+4.5.6+5.6.7}{3.3.2.3.3.1+2.3.3.3.4.3+3.3.4.3.3.5+3.4.5.3.6.3+3.5.3.6.7.3}+\frac{8}{27}\)
\(C=\frac{1.2.3+2.3.4+3.4.5+4.5.6+5.6.7}{3^3.\left(1.2.3+2.3.4+3.4.5+4.5.6+5.6.7\right)}+\frac{8}{27}\)
\(C=\frac{1}{3^3}+\frac{8}{27}=\frac{1}{27}+\frac{8}{27}=\frac{9}{27}=\frac{1}{3}\)
Vậy C = \(\frac{1}{3}\)
Đặt C =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow C=\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\div2\)
A = 1.2.3+23.1.2.3+33.1.2.3+43.1.2.3 /3.4.5+23.3.4.5+33.3.4.5+43.3.4.5
=1.2.3 (1+23+33+43)/3.4.5 (1+23+33+43)
=1.2.3/3.4.5
=1/10
B=10101 ( 15/333333 + 2/333333 +57 / 333333 )
B=10101 . 74 /333333
=74/33
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