s=1/3+1/9+1/27+....1/729
Câu 3. (2 điểm) Tính nhanh tổng sau
S = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) +\(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)+ \(\dfrac{1}{729}\)
S= 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
S= 3 x ( 1/1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 )
S = 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
S= 3 +1 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 - 1 - 1/9 -1/27 - 1/81 - 1/243 - 1/729
S = 3 - 1/729
S= 142/729
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Các bạn giúp mình với
\(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times S=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times S-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)
\(2\times S=3-\frac{1}{729}\)
\(S=\frac{1093}{729}\)
1/3+1/9+1/27+1/81+1/243+1/729
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+1/3+1/9+1/27+1/81+1/243+1/729
\(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}\)+\(\dfrac{1}{243}+\dfrac{1}{729}\)=\(\dfrac{1093}{729}\)
\(=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)
\(\dfrac{729}{729}+\dfrac{243}{729}+\dfrac{81}{729}+\dfrac{27}{729}+\dfrac{9}{729}+\dfrac{3}{729}+\dfrac{1}{729}\)
\(=\dfrac{\left(729+243+81+27+9+3+1\right)}{729}=\dfrac{1084}{729}\)
1+1/3+1/9+1/27+1/81+1/243+1/729
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\\ \Rightarrow A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\\ \Rightarrow\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}\\ \Rightarrow\dfrac{1}{3}A-A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^7}-1-\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^6}\\ \Rightarrow-\dfrac{2}{3}A=\dfrac{1}{3^7}-1\\ \Rightarrow A=\left(\dfrac{1}{2187}-1\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\left(-\dfrac{2186}{2187}\right):\left(-\dfrac{2}{3}\right)\\ \Rightarrow A=\dfrac{1093}{729}\)
Các bạn làm như vậy với các cháu học sinh lớp 4, 5 là ko làm đc. KQ tính bằng 1093/729 là đúng nhưng PP làm chưa đúng.
Mình hướng dẫn con mình làm như thế này là phù hợp với kiến thức lớp 4:
Ta tách phân số như sau:
= (5/3-2/3) + (2/3-1/3) + (1/3-2/9) + (2/9-5/27) + (5/27-14/81) + (14/81-41/243) + (41/243-122/729)
Sau khi rút gọn ta còn:
= 5/3 - 122/729
= (5*243-122)/729
= 1093/729
1+1/3+1/9+1/27+1/81+1/243+1/729
A = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\) + \(\dfrac{1}{729}\)
3 \(\times\) A = 3 + 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\) + \(\dfrac{1}{243}\)
3 \(\times\) A - A = 3 - \(\dfrac{1}{729}\)
A \(\times\)(3-1) = \(\dfrac{2186}{729}\)
A \(\times\) 2 = \(\dfrac{2186}{729}\)
A = \(\dfrac{2186}{729}\): 2
A = \(\dfrac{1093}{729}\)
B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 +1/729
B=243/729+81/729+27/727+9/727+3/727+1/727=364/727
B=243/729+81/729+27/727+9/727+3/727+1/727=364/727
1/3 + 1/9 + 1/27 + 1/81 + 1/24 + 1/729
Đặt A=1/3 + 1/9 + 1/27 + 1/81 + 1/24 + 1/729
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^6}\)
\(3A=3\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^5}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(2A=1-\frac{1}{3^6}\)
\(A=\frac{1-\frac{1}{3^6}}{2}\)
hâm ak giải cách tiểu học cho tui dễ hiểu !!!
Lê Xuân Hoan:ủa lớp 6 học lũy thừa rồi mà