F = 3 + 9 + 27 + 81 + .... + 6561 + 19.683
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M = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
=> 3M = \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\)
=> 3M - M = ( \(1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{2187}\) ) - ( \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\))
2M = 1 - \(\frac{1}{6561}\)
2M = \(\frac{6560}{6561}\)
=> M = \(\frac{3280}{6561}\)
\(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.......+\frac{1}{6561}\)
\(\Rightarrow M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\)
\(\Rightarrow3M=3\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+.........+\frac{1}{3^8}\right)\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+............+\frac{1}{3^7}\)
\(\Rightarrow3M-M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..........+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-.......-\frac{1}{3^8}\)
\(\Rightarrow2M=1-\frac{1}{3^8}\)
\(\Rightarrow M=\frac{1-\frac{1}{3^8}}{2}\)
Vậy M = \(\frac{1-\frac{1}{3^8}}{2}\)
3M=1+1/3+1/9+...+1/2187
2M=3M-M
2M=1-1/6561
2M=6560/6561
M=3280/6561
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
Ta có:
\(S=3.2^0-3^1+3.2^1-3^2+3.2^2+3.2^3-3^3+3.2^4-3^4+...-3^7+3.2^{10}+3.2^{11}-3^8+3.2^{12}\)
\(=3.\left(2^0+2^1+2^2+2^3+2^4+...+2^{10}+2^{11}+2^{12}\right)-\left(3^1+3^2+3^3+...+3^7+3^8\right)\)
Đặt: \(A=2^0+2^1+2^2+...+2^{11}+2^{12}\)
=> \(2.A=2^1+2^2+2^3+...+2^{12}+2^{13}\)
=> \(2.A-A=2^{13}-2^0\)
\(\Rightarrow A=2^{13}-1=8191\)
Đặt: \(B=3^1+3^2+3^3+...+3^8\)
\(\Rightarrow3.B=3^2+3^3+3^4+...+3^9\)
=> \(3B-B=3^9-3^1=19680\)
=> \(2B=19680\Rightarrow B=9840\)
=> S=3.A-B=3.8191-9840=14733
Cho \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(\frac{1}{3}A=\frac{1}{3}\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\right)\)
\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{19683}\)
\(A-\frac{1}{3}A=\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right)-\left(\frac{1}{9}+\frac{1}{27}+...+\frac{1}{19683}\right)\)
\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{19683}\)
\(A=\frac{4840}{9683}:\frac{2}{3}=\frac{7260}{9683}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=3.\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\) \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)
\(\Rightarrow2A=1-\frac{1}{3^8}\) \(\Rightarrow A=\frac{1-\frac{1}{3^8}}{2}\)
k cho mik đi mn!Nguyễn Như Quỳnh!
đặt A = 1+3+9+27+....+2187+6561
=>A = 30 + 31 + 32 + 33 + .. . +37 + 38
3A = 31 + 32 + 33 + ... + 38 + 39
3A - A = (31 + 32 + 33 + ... + 38 + 39)-(30 + 31 + 32 + 33 + .. . +37 + 38 )
2A = 39 - 1
A=\(\frac{3^9-1}{2}=\frac{19682}{2}=9841\)