Tính hợp lý tổng sau
S=1+1/3+1/9+1/27+....+1/2187
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HD
1
30 tháng 3 2019
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}\)
\(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(3A-A=\left[1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right]-\left[\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right]\)
\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)
\(A=\frac{6560}{6561}:2\)
\(A=\frac{3280}{6561}\)
Vậy : ...
OD
3
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)
\(3B=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{6561}\right)\)
\(3B=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)
\(2B=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+...+\left(1-\dfrac{1}{6561}\right)\)
\(2B=0+0+...+1-\dfrac{1}{6561}\)
\(2B=1-\dfrac{1}{6561}\)
\(B=\left(1-\dfrac{1}{6561}\right):2\)
\(B=\dfrac{6560}{6561}:2\)
\(B=\dfrac{3280}{6561}\)
PT
4
KT
26 tháng 7 2018
\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow\)\(3S=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(\Rightarrow\)\(3S-S=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(\Rightarrow\)\(2S=1-\frac{1}{3^7}\)
\(\Rightarrow\)\(S=\frac{1-\frac{1}{3^7}}{2}\)
26 tháng 7 2018
\(S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(3S=1+\frac{1}{3}+...+\frac{1}{3^6}\)
\(3S-S=\left(1+\frac{1}{3}+...+\frac{1}{3^6}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(2S=1-\frac{1}{3^7}\)
\(S=\frac{1-\frac{1}{3^7}}{2}\)
PH
2
NC
27 tháng 8 2020
Bài làm:
Ta có: \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)
=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
=> \(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
<=> \(2A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=> \(A=\frac{3^8-1}{3^8.2}\)
B
27 tháng 8 2020
Bài làm :
Ta có :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(\Rightarrow3\times A=\frac{1\times3}{3}+\frac{1\times3}{9}+\frac{1\times3}{27}+...+\frac{1\times3}{6561}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\left(\frac{1}{6561}-\frac{1}{6561}\right)\)
\(3\times A=1+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\frac{1}{6561}\right)-\frac{1}{6561}\)
\(3\times A=1+A-\frac{1}{6561}\)
\(\Rightarrow2\times A=1-\frac{1}{6561}\)( Trừ bỏ A ở cả 2 vế )
\(2\times A=\frac{6560}{6561}\)
\(A=\frac{6560}{6561}\div2=\frac{3280}{6561}\)
Vậy A=3280/6561
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
LT
2
13 tháng 7 2016
P = 78 × 31 + 78 × 24 + 78 × 17 + 22 × 72
P = 78 × (31 + 24 + 17) + 22 × 72
P = 78 × 72 + 22 × 72
P = 72 × (78 + 22)
P = 72 × 100
P = 7200
S = 1 + 1/3 + 1/9 + 1/27 + ... + 1/2187
3S = 3 + 1 + 1/3 + 1/9 + ... + 1/729
3S - S = (3 + 1 + 1/3 + 1/9 + ... + 1/729) - (1 + 1/3 + 1/9 + 1/27 + ... + 1/2187)
2S = 3 - 1/2187
2S = 6560/2187
S = 6560/2187 : 2
S = 6560/2187 × 1/2
S = 3280/2187
C
0
TD
1
14 tháng 3 2023
\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)
\(3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}\)
Lấy 3A - A ta được :
\(2A=1-\dfrac{1}{6561}=\dfrac{6560}{6561}\Leftrightarrow A=\dfrac{6560}{6561}:2\)
\(\Leftrightarrow A=\dfrac{6560}{6561}.\dfrac{1}{2}=\dfrac{3280}{6561}\)
NH
1
TC
0
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
\(2S=3-\frac{1}{3^7}\)
\(S=\frac{3-\frac{1}{3^7}}{2}\)
S= 1+ \(\frac{1}{3}\)+ \(\frac{1}{9}\)+...+ \(\frac{1}{729}\)+ \(\frac{1}{2187}\).
=> S= 1+ \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{3^6}\)+ \(\frac{1}{3^7}\).
=>3S= 3+ 1+ \(\frac{1}{3}\)+...+ \(\frac{1}{3^5}\)+ \(\frac{1}{3^6}\).
=> 3S- S=( 3+ 1+ \(\frac{1}{3}\)+...+ \(\frac{1}{3^5}\)+ \(\frac{1}{3^6}\))-( 1+ \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+...+ \(\frac{1}{3^6}\)+ \(\frac{1}{3^7}\)).
=> 2S= 3- \(\frac{1}{3^7}\).
=> 2S= 3- \(\frac{1}{2187}\).
=> 2S= \(\frac{6560}{2187}\).
=> S= \(\frac{6560}{2187}\): 2.
=> S= \(\frac{3280}{2187}\).
Vậy S= \(\frac{3280}{2187}\).