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TT
0
XO
30 tháng 1 2022
\(P=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}=\frac{49}{99}\)
NT
2
10 tháng 4 2016
B=1.3+2.4+3.5+...+97.99+98.100B=1.3+2.4+3.5+...+97.99+98.100
B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)B=1(2+1)+2(3+1)+....+97(98+1)+98(99+1)
B=1.2+1+2.3+2+....+97.98+97+98.99+98B=1.2+1+2.3+2+....+97.98+97+98.99+98
B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)B=(1.2+2.3+3.4+....+97.98+98.99)+(1+2+3+...+98)
B=98.99.1003+98.992B=98.99.1003+98.992
B=323400+4851=328251B=323400+4851=328251
10 tháng 4 2016
Số đó=1.3 + 2.4 + 3.5 +....+ 98.100
= 1(2+1) + 2.(3+1) + 3.(4+1) +...+ 98(99+1)
= 1.2 + 1 + 2.3 + 2 + 3.4 + 3+....+ 98.99 +98
= (1.2 + 2.3 + 3.4+....98.99) + (1+2+3+....+98)
=323400 + 4851=328251
LD
1
23 tháng 7 2015
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}\)
\(=\frac{3}{7}\)
b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
1 tháng 7 2015
\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{97\times99}\)
\(=\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)+...+\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{99-3}{297}\)
\(=\frac{96}{297}=\frac{32}{99}\)
8 tháng 7 2015
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\)
\(=\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{97}-\frac{2}{99}\right):2\)
\(=\left(2-\frac{2}{99}\right):2=\frac{98}{99}\)
8 tháng 7 2015
D = 1 - 1/3 + 1/3 - 1/5 + .... + 1/97 - 1/99
D = 1 - 1/99
D = 98/99
10 tháng 8 2015
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:2=\frac{100}{101}\times\frac{1}{2}=\frac{50}{101}\)
\(A=1.3+3.5+5.7+...+97.99\)
\(\Rightarrow6A=1.3.6+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+97+99.\left(101-95\right)\)
\(\Rightarrow6A=1.3.6+3.5.7-1.3.5+5.7.9-3.5.7+...+97.99.101-95.97.99\)
\(\Rightarrow6A=1.3.6+97.99.101-1.3.5\)
\(\Rightarrow6A=3.\left(1+97.33.101\right)\)
\(\Rightarrow2A=1+323301\)
\(\Rightarrow2A=323302\)
\(\Rightarrow A=161651\)