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Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
a,1+1+2+2+3+3+...+100+100
=1x2+2x2+3x2+...+100x2
=2x(1+2+3+...+100)
=\(2.\frac{\left(100+1\right).\left[\left(100-1\right):1+1\right]}{2}\)
=2x5050
=10100
Chú ý dấu . là x
a, 1+1+2+2+3+...+100+100
=1+2+3+...+100+1+2+3+...+100
= (100+1)*50 /2 + (100+1)*50/2
=5050+5050
=11000
\(x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(x=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
\(\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}+\frac{1}{10x11}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{6}-\frac{1}{11}\)
\(\frac{5}{66}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-......-\left(\frac{1}{6}-\frac{1}{6}\right)-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
1/2+1/6+1/12+1/20+1/30+1/42
=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
b) 1/3+1/3^2+1/3^3+1/3^4+1/3^5 (goi tong bang M)
3M=1+1/3+1/3^2+1/3^3+1/3^4
3M-M=1-1/3^5
2M=242/243
M=242/243*1/2=121/243
mình gọi phép tính trên là Acho đỡ rối nha
có A=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
A=1-1/32
A=31/32
Ta có:
1/2+1/4+1/8+1/16+1/32
=16/32+8/32+4/32+2/32+1/32
=31/32
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{21111111}-\frac{2}{21111112}\)
= \(\frac{1}{1}-\frac{1}{21111112}\)(\(-\frac{1}{2}\)rút gọn cho \(+\frac{1}{2}\)và cứ như vậy đến khi chỉ còn 2 phân số \(\frac{1}{1}\)và \(\frac{1}{21111112}\))
= \(\frac{21111111}{21111112}\)
100% đúng nha bạn
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{201111111}-\frac{1}{2011111112}\)
\(\frac{1}{1}-\frac{1}{201111112}\)
\(\frac{201111111}{201111112}\)