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NT
1
AH
Akai Haruma
Giáo viên
28 tháng 9
Lời giải:
Đặt $A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}$
$3\times A=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}$
$3\times A-A=3-\frac{1}{2187}$
$2\times A=3-\frac{1}{2187}=\frac{6560}{2187}$
$A=\frac{6560}{2187}:2=\frac{3280}{2187}$
PH
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 6 2023
(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))
(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))
Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)
Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2
Dãy số trên có số số hạng là: (25 - 1): 2 + 1 = 13
Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)
A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))
A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)( \(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))
A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)
A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)
Đặt B = \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\)
B\(\times\)3 =1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)
B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)
2B = \(\dfrac{242}{243}\)
B = \(\dfrac{242}{243}\): 2
B = \(\dfrac{121}{243}\)
11a + B = 11a + \(\dfrac{121}{243}\) (2)
Từ (1) và(2) ta có:
a\(\times\)13 + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)
a \(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\)
\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)
a \(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)
a \(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)
a \(\times\) 2 = \(\dfrac{109}{6075}\)
a = \(\dfrac{109}{6075}\): 2
a = \(\dfrac{109}{12150}\)
PN
1
16 tháng 8 2017
Câu trả lời hay nhất: Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x ﴾1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729﴿
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
A x 3 ‐ A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 ‐ ﴾1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729﴿
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 ‐ 1/3 ‐ 1/9 ‐ 1/27 ‐ 1/81 ‐ 1/243 ‐ 1/729
= 1 ‐ 1/729
A x 2 = 728/729
A = 364/729
NHỚ TK MK NHA
C
2
9 tháng 6 2023
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
\(3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
\(3A-A=\left(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\right)\)
\(2A=3-\dfrac{1}{729}=\dfrac{2186}{729}\)
\(A=\dfrac{2186}{729}\div2=\dfrac{1093}{729}\)
9 tháng 6 2023
A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
3A = \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)
3A - A = ( \(3+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )
2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)
A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)
HT
2
15 tháng 7 2015
A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049
3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
- ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
= 1 - 1/59049
2 x A = 59048/59049
A = 59048/59049 : 2
A = 29524/59049
21 tháng 8 2019
A = 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049
3 x A = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
3 x A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683
3 x A - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/19683 - ( 1 + 1/3 + 1/9 + 1/27 + 1/81 + ... + 1/59049 )
= 1 - 1/59049
2 x A = 59048/59049
A = 59048/59049 : 2
A = 29524/59049
LT
3
31 tháng 8 2019
Ta có:\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
Xét\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Leftrightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{729}\)
\(\Leftrightarrow\frac{2}{3}A=\frac{243-1}{729}\Leftrightarrow A=\frac{3}{2}\times\frac{242}{729}=\frac{121}{243}\)
31 tháng 8 2019
Phải là : A=1/3+1/9+1/27+1/81+1/243 ta có: 3A=1+1/3+1/9+1/27+1/81 3A-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243)=1-1/243 2A=242/243 A=242/243:2=121/243
Dấu "." là dấu nhân bạn nhé.
Ta có:
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{729.3}\)
\(\Rightarrow3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\)
\(\Rightarrow3A-A=3-\dfrac{1}{729.3}\)
\(\Rightarrow2A=3-\dfrac{1}{729.3}\)
\(\Rightarrow A=\dfrac{1}{2}\left(3-\dfrac{1}{729.3}\right)\)