Tính tổng sau đây bằng cách thuận tiện nhất :
` S = 1/3 + 1/9 + 1/27 + 1/81 `
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Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
\(3A=3+1+...+\frac{1}{3^4}\)
\(3A-A=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2A=3-\frac{1}{3^5}\)
\(A=\frac{3-\frac{1}{3^5}}{2}\)
Đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(S=1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\)
\(S\times3=\left(1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\right)\times3\)
\(S\times3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Xét: \(S\times3-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(S\times2=3-\frac{1}{243}\)
\(S\times2=\frac{728}{243}\)
\(S=\frac{728}{243}\div2\)
\(S=\frac{364}{243}\)
Vậy \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)\)
\(2\times A=1-\frac{1}{729}=\frac{728}{729}\)
\(A=\frac{364}{729}\)
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 = 3 +1+ 1/3 + 1/9 + 1/27 + 1/81
Vậy:
S x 3 - S = 3 - 1/243
2S = 728/243
S = 364/243
tick đúng nha
= 27/81 + 9/81 + 3/81 + 1/81
= ( 27/81 + 3/81 ) + ( 9/81 + 1/81 )
= 30/81 + 10/81
= 40/81
# Math is easy
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)
\(A=1-\frac{1}{64}\)
\(A=\frac{63}{64}\)
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(3B-B=1-\frac{1}{243}\)
\(2B=\frac{242}{243}\)
\(B=\frac{242}{243}\div2\)
\(B=\frac{121}{243}\)
a.A=1/2+1/4+1/8+1/16+1/32+1/64
A= \(\frac{1}{1\cdot2}+\frac{1}{2\cdot2}+\frac{1}{2\cdot4}+\frac{1}{4\cdot4}+\frac{1}{4\cdot8}+\frac{1}{8\cdot8}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{8}\)
= 1 - 1/8 = 7/8
b.B=1/3+1/9+1/27+1/81+1/243
B= \(\frac{1}{1\cdot3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot9}+\frac{1}{9\cdot9}+\frac{1}{9\cdot27}\)
= 1 - 1/27 = 26/27
Số số hạng : ( 49 - 1 ) : 2 +1 = 25
Tổng : ( 49 + 1 ) x 25 : 2 = 625
S = 1/3+1/9+1/27+1/81+1/243+1/729+1/2187 ( 1 )
Nhân S với 3. Ta có:
S x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 ( 2 )
Trừ ( 2 ) với ( 1 ) ta có:
S x 3 - S = 1 - 1/ 2187
2S = 2186/ 2187
S = 2186/ 2187 : 2
S = 1093/ 2187
\(\frac{9}{10}+\frac{7}{9}+\frac{5}{8}+\frac{3}{7}+\frac{3}{5}+\frac{2}{5}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}\)
= \(\left(\frac{9}{10}+\frac{1}{10}\right)+\left(\frac{7}{9}+\frac{2}{9}\right)+\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
= \(\frac{10}{10}+\frac{9}{9}+\frac{8}{8}+\frac{7}{7}+\frac{5}{5}\)
= \(1+1+1+1+1\)
= \(1\times5\)
= \(5\)
Gọi A là tổng của 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10, ta có :
A = 9/10 + 7/9 + 5/8 + 3/7 + 3/5 + 2/5 + 4/7 + 3/8 + 2/9 + 1/10
A = (9/10 + 1/10) + (7/9 + 2/9) + (5/8 + 3/8) + (3/7 + 4/7) + (3/5 + 2/5)
A = 1 + 1 + 1 + 1 + 1
A = 5
= 1 x 27/3x27 + 1x9/9x9 + 1x3 / 27 x 3 + 1/81
=27/81 + 9/81 + 3/81 + 1/81
= 40/81
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