B = 2 + 2/3 + 2/9 + 2/27 + .... + 2/279 + 2/ 2187
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\(A=\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\)
\(3\times A=2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{729}+\frac{2}{2187}\)
\(3\times A-A=\left(2+\frac{2}{3}+\frac{2}{9}+...+\frac{2}{2187}\right)-\left(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{2187}+\frac{2}{6561}\right)\)
\(2\times A=2-\frac{2}{6561}\)
\(A=\frac{6560}{6561}\)
a) 48 : 6 : 2 = 4
48 : 6 x 2 = 16
b) 27 : 9 x 3 = 9
27 : 9 : 3 = 1
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 ~
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\(B=2.\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(3.B=2.\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)\)
\(3B-B=2\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^7}\right)\)
\(B=1-\frac{1}{3^7}\)
P/s dấu "." này là dấu x nha bn !
_Kudo_