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\(M=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{6.7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}=1-\dfrac{1}{7}=\dfrac{6}{7}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{6\times7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}=1-\frac{1}{7}=\frac{6}{7}\)
\(\frac{1}{2} +\frac{1}{6}+\frac{1}{21}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{23}{28}\)
Gọi tổng trên là A
A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7
A = 1 - 1/2 + 1/2 - 1/3 +.......+ 1/6 - 1/7
A = 1 - 1/7
A = 6/7
1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7=7-1/7=6/7
1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2
= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7
= 1 - 1/7
= 6/7
Chú ý : Dấu " . " là dấu x
= 1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7
= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
= 1-1/7
= 6/7
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42
= 2/3 + 1/12 + 1/20 + 1/30 + 1/42
= 3/4 + 1/20 + 1/30 + 1/42
= 4/5 + 1/30 + 1/42
= 5/6 + 1/42
= 6/7
đ/s : ...
A=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
A=1-1/7
A=6/7
A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7
A = 1/1 - 1/2 + 1/2 - 1/3 + 1 /3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7
A = 1/1 - 1/7
A = 6/7
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}=\frac{6}{7}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{45}{28}\)
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{10\times11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
Đặt A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
=\(\frac{10}{11}\)
Vậy...
=(1/1-1/2)+(1/2-1/3)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7) =(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+(1/5-1/5)+(1/6-1/6)+(1/1-1/7) = 0+0+0+0+0+6/7 =6/7