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Ta có G = 2 . 4 + 4 . 6 + 6 . 8 + ... + 98 . 100
6 . G = 2 . 4 . 6 + 4 . 6 . 6 + 6 . 8 . 6 + .... + 98 . 100 . 6
6 . G = 2 . 4 . 6 + 4 . 6 . ( 8 - 2 ) + 6 . 8 . ( 10 - 4 ) + .... + 98 . 100 . ( 102 - 96 )
6 . G = 2 . 4 . 6 + 4 . 6 . 8 - 2 . 4 . 6 + 6 . 8 . 10 - 4 . 6 . 8 + ... + 98 .100 . 102 - 96 . 98 . 100
6 . G = 98 . 100 . 102
G = 98 . 100 . 102 : 6
G = 999 600 : 6
G = 166 600
Ta có G = 2 . 4 + 4 . 6 + 6 . 8 + ... + 98 . 100
6 . G = 2 . 4 . 6 + 4 . 6 . 6 + 6 . 8 . 6 + .... + 98 . 100 . 6
6 . G = 2 . 4 . 6 + 4 . 6 . ( 8 - 2 ) + 6 . 8 . ( 10 - 4 ) + .... + 98 . 100 . ( 102 - 96 )
6 . G = 2 . 4 . 6 + 4 . 6 . 8 - 2 . 4 . 6 + 6 . 8 . 10 - 4 . 6 . 8 + ... + 98 .100 . 102 - 96 . 98 . 100
6 . G = 98 . 100 . 102
G = 98 . 100 . 102 : 6
G = 999 600 : 6
G = 166 600
\(A=\frac{2.6.10+6.10.14+...+1994.1998.2002}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{1.2^3.3.5+3.2^3.5.7+....+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=8\)
Vậy A = 8
\(\dfrac{1\times2\times3+2\times4\times6+4\times8\times12+7\times14\times21}{1\times3\times5+2\times6\times10+4\times12\times20+7\times21\times35}\\ =\dfrac{1\times2\times3}{1\times3\times5}+\dfrac{2\times4\times6}{2\times6\times10}+\dfrac{4\times8\times12}{4\times12\times20}+\dfrac{7\times14\times21}{7\times21\times35}\\ =\dfrac{2}{5}+\dfrac{4}{10}+\dfrac{8}{20}+\dfrac{14}{35}\\ =\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}\\ =\dfrac{2}{5}\times4\\ =\dfrac{8}{5}\)
