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\(A=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\)
\(3\times A=2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{729}+\frac{2}{2187}\)
\(3\times A-A=\left(2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{2187}\right)-\left(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\right)\)
\(2\times A=2-\frac{2}{6561}\)
\(A=\frac{6560}{6561}\)
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
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
A= 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 + 1/2187
Nhân A với 3 , ta có :
A x 3 = 3x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 + 1/2187)
A x 3 = 1+ ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
A x 3 = 1 + A - 1/2187
A x 2 = 1 - 1/2187
A x 2 = 2186 / 2187
A = 2186 / 2187 : 2
A = 1093/2187
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+...+\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}\)
\(\Rightarrow\)\(A=\frac{2047}{2048}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B-B=1-\frac{1}{2187}\)
\(2B=\frac{2186}{2187}\)
\(\Rightarrow B=\frac{2186}{4374}=\frac{1093}{2187}\)
grydew5ewry63h đh djen djjjdhe ụgejgetewittrtyejkteytwthtyuiegejfv ehrttd hdtwwnv
\(B=2.\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(3.B=2.\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(3B-B=2\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^7}\right)\)
\(B=1-\frac{1}{3^7}\)
P/s dấu "." này là dấu x nha bn !
_Kudo_