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lấy MS chung là 2187, ta có:
729 + 243 + 81 + 9 + 3 + 1
________________________ = 1066/2187
2187
ta có :
= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )
= 59050 + 2190 + 6570 + 270 + 810
= 59050 + ( 2190 + 810 ) + 6570 + 270
= 59050 + 3000 + 6570 + 270
= 59050 + ( 3000 + 6570 ) + 270
= 59050 + 9570 + 270
= 68620 + 270
= 68890
Đặt \(V=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(\Rightarrow3V=3.\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}+\dfrac{1}{2187}\right)\)
\(\Rightarrow3V=1+\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+...+\dfrac{1}{729}\right)\)
\(\Rightarrow3V=1+V-\dfrac{1}{2187}\)
\(\Rightarrow2V=1-\dfrac{1}{2187}\)
\(\Rightarrow V=\dfrac{1093}{2187}\).
A = 1/3 + 1/9 + 1/27 + 1/81 +...+1/729 + 1/2187
3A = 1 + 1/3 + 1/9 + 1/27 + 1/81 +...+1/729
=>2A = 1 - 1/2187
=> A = ....
Gọi tong trên là A
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)
\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)
\(2A=1-\frac{1}{2187}\)
\(2A=\frac{2186}{2187}\)
\(A=\frac{2186}{2187}:2\)
\(A=\frac{1093}{2187}\)
Vậy tổng A = \(\frac{1093}{2187}\)
\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)
\(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)
=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)
<=> 2y = 3- 1/2187
=> y = \(\frac{3-\frac{1}{2187}}{2}\)
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+...+\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}\)
\(\Rightarrow\)\(A=\frac{2047}{2048}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B-B=1-\frac{1}{2187}\)
\(2B=\frac{2186}{2187}\)
\(\Rightarrow B=\frac{2186}{4374}=\frac{1093}{2187}\)
-597871
nhầm nha