1/3+2/9+3/27+4/81+...+7/2187 = ?
1/3 + 1/9 + 1/27 + 1/81 +...+1/729 + 1/2187
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
=> 3A - A = 1 - 1/2187
=> 2A = ... => A = ...1/3 +1/9 + 1/27 + 1/81 + 1/243 + 1/729 + 1/2187 =?
lấy MS chung là 2187, ta có:
729 + 243 + 81 + 9 + 3 + 1
________________________ = 1066/2187
2187
1066/2187.
tính nhanh 1+3+9+27+81+243+729+2187+6561+19683+59049
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
Kết quả là 68890
Nhớ trả lời cho mình
h = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/2187 ai giúp mình với
Ta có \(H=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+....+\frac{1}{2187}\)
Suy ra \(3H=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{729}\)
Nên \(3H-H=1-\frac{1}{2187}\)hay \(2H=1-\frac{1}{2187}\)
Do đó \(H=\frac{1}{2}-\frac{1}{4374}=\frac{1093}{2187}\)
Vậy \(H=\frac{1093}{2187}\)
Tính:\(S=3-3+6-9+12+24-27+48-81+...-2187+3072+6144-6561+12288\)
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
Tính nhanh tổng sau
A=1/6+1/12+1/20+...1/9900
B=1/3+1/9+1/27+1/81+...=1/2187
1.Có bao nhiêu giá trị của \(\left|2x-1\right|=5\) thỏa mãn
A. 0 B. 1 C. 3 D.2
2. Tìm x, biết \(\dfrac{\left(-3\right)^x}{81}=-27\)
A. x=7 B. x=2187 C. x=3 D. x = -7
\(1,\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\1-2x=5\end{matrix}\right.\Leftrightarrow D\\ 2,\Leftrightarrow\left(-3\right)^x=-27\cdot81=-2187=\left(-3\right)^7\\ \Leftrightarrow x=7\left(A\right)\)
Tính tổng:
C=1+2+4+8+...+1024
D=1+3+9+27+...+2187
C = 1 + 2 + 4 + 8 + ... + 1024
2 x C = 2 + 4 + 8 + ... + 1024 + 2048
2 x C - C = C = (2 + 4 + 8 + ... + 1024 + 2048) - (1 + 2 + 4 + 8 + ... + 1024) = 2048 - 1 = 2047
D = 1 + 3 + 9 + 27 + ... + 2187
3 x D = 3 + 9 + 27 + ... + 2187 + 6561
3 x D - D = 2 x D = (3 + 9 + 27 + ... + 2187 + 6561) - (1 + 3 + 9 + 27 + ... + 2187) = 6561 - 1 = 6560
D = 6560 : 2 = 3280
Tính nhanh \(\frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
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 = \(\frac{3-\frac{1}{2187}}{2}\)
\(\text{Đ}\text{ặt} A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(\Rightarrow2187A=2187+729+243+81+27+9+3+1\)
\(\Leftrightarrow2187A=3280\)
\(\Leftrightarrow A=\frac{3280}{2187}\)
Chắc chắn 100% luôn