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3xB=3x(1/4+1/12+1/36+1/108+1/324+1/972+1/2916+1/8748)
3xB=3/4 + 1/4 +1/12 +1/36 +.........+1/2916
3xB - B= (3/4 + 1/4 + 1/12+1/36 + .........+1/2916) - ( 1/4 +1/12 +1/36 +1/108 + 1/324 + 1/972 + 1/2916 +1/8748 )
2xB =3/4 - 1/8748
2xB =1640/2187
B = 1640/2187 :2
B = 820/2187.
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}.\)
\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right).\)
\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{972}\)
\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
\(2A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}-\frac{1}{4}-\frac{1}{12}-\frac{1}{36}-\frac{1}{108}-\frac{1}{324}-\frac{1}{972}\)
\(2A=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)
\(\Rightarrow A=\frac{182}{243}:2=\frac{91}{243}\)
\(\Rightarrow A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}=\frac{91}{243}\)
ta có
A x 3 =3/4 + 3/12 + 3/36 +3/108 + 3/324 +3/972
A x 3=3/4 +1/4 + 1/12 +1/36 +1/108
A x 3 - A =(3/4 +1/12+1/36 +1/108)-(1/4 +1/12 +1/36 +1/108 +1/324 + 1/972)
A x 2=3/4 +1/12 +1/36 +1/108 - 1/4 -1/12 -1/36-1/108 -1/324 -1/972
A x 2= 3/4 - 1/972
A x 2= 728/972
A =728/972 : 2
A=91/243
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)
=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(=1-\frac{1}{2^9}=\frac{511}{512}\)
\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)
=>\(3B=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
=>\(3B-B=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}\right)\)
=>\(2B=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)
=>\(B=\frac{182}{243}:2=\frac{91}{243}\)
\(3C=1+\frac{1}{3}+.....+\frac{1}{3^{2015}}\)
\(\Rightarrow3C-C=2C=\left(1+\frac{1}{3}+.....+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}.....+\frac{1}{3^{2015}}\right)=1-\frac{1}{3^{2015}}\)
\(\Rightarrow C=\frac{3^{2015}-1}{3^{2015}.2}\)
\(D=4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)
\(\Rightarrow3D=4\left(3+1+....+\frac{1}{3^4}\right)\)
\(\Rightarrow3D-D=2D=4\left(3+1+....+\frac{1}{3^4}\right)-4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)
\(\Rightarrow2D=4\left(3-\frac{1}{3^5}\right)\Rightarrow D=2\left(3-\frac{1}{3^5}\right)\)
\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=1-\frac{1}{2^9}\)
\(A=\frac{2^9-1}{2^9}\)
Vậy \(A=\frac{2^9-1}{2^9}\)
Chúc bạn học tốt ~
a) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right):2\)
\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=5.\left(1-\frac{1}{101}\right):2=5.\frac{100}{101}:2=\frac{500}{101}.\frac{1}{2}\)\(=\frac{250}{101}\)
b) \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)\(.\frac{1}{3}\)
\(=(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}).\frac{1}{3}\)
\(=(\frac{1}{3}-\frac{1}{33}).\frac{1}{3}=\frac{10}{33}.\frac{1}{3}=\frac{10}{99}\)
91/243
91/243 đó em