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Đặt A= 1/4+1/12+1/36+1/108+1/324 +1/972
=1/4.(1+1/3+1/9+1/27+1/81+1/243)
=1/4(1+1/3 +1/3^2 +1/3^3 +1/3^4 +1/3^5)
gọi M=1+1/3+1/3^2+1/3^3+1/3^4+1/3^5
3M=3+1+1/3+1/3^2+1/3^3+1/3^4
3M-M=(3+1+1/3+1/3^2+1/3^3+1/3^4)-(1+1/3+1/3^2+1/3^3+1/3^4+1/3^5)
2M=3-1/3^5
M=3/2-1/3^5.2
A=1/4.(3/2-1/3^5.2)
A=3/8-1/1944=91/243
Vậy .......
Chúc bạn học tốt .....^-^
Đặt tổng trên là A ta có
A/3=1/12+1/36+1/108+ +1/2916+1/8748
A-A/3=2A/3=
=1/4-1/8748=2186/8748
=>A=1093/1916
k mk nhá
Mk đã giúp bn rồi đấy
Ai k mk ,mk k lại,cảm ơn các bn
HC TỐT NHÁ BN
#HYISTHEBEST#
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)
=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)
\(=1-\frac{1}{2^9}=\frac{511}{512}\)
\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)
=>\(3B=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)
=>\(3B-B=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}\right)\)
=>\(2B=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)
=>\(B=\frac{182}{243}:2=\frac{91}{243}\)
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