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\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{2187}+\dfrac{1}{6561}\)
\(3B=3\cdot\left(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{6561}\right)\)
\(3B=1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{2187}\right)-\left(\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{6561}\right)\)
\(2B=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+...+\left(1-\dfrac{1}{6561}\right)\)
\(2B=0+0+...+1-\dfrac{1}{6561}\)
\(2B=1-\dfrac{1}{6561}\)
\(B=\left(1-\dfrac{1}{6561}\right):2\)
\(B=\dfrac{6560}{6561}:2\)
\(B=\dfrac{3280}{6561}\)
ta có :
= ( 1 + 59049 ) + ( 3 + 2187 ) + ( 9 + 6561 ) + ( 27 + 243 ) + ( 81 + 729 )
= 59050 + 2190 + 6570 + 270 + 810
= 59050 + ( 2190 + 810 ) + 6570 + 270
= 59050 + 3000 + 6570 + 270
= 59050 + ( 3000 + 6570 ) + 270
= 59050 + 9570 + 270
= 68620 + 270
= 68890
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2048}\)
\(A=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+...+\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(A=1-\frac{1}{2048}\)
\(\Rightarrow\)\(A=\frac{2047}{2048}\)
\(3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3B-B=1-\frac{1}{2187}\)
\(2B=\frac{2186}{2187}\)
\(\Rightarrow B=\frac{2186}{4374}=\frac{1093}{2187}\)
Tham khảo tại đây
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
Chúc học tốt!
Đặt \(B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(\Rightarrow3B=3.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right)\)
\(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\)
\(\Rightarrow3B-B=\left(3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)-\)\(\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right)\)
\(\Rightarrow2B=3-\frac{1}{2187}\)
\(\Rightarrow B=\left(3-\frac{1}{2187}\right):2\)
\(\Rightarrow B=\frac{6560}{2187}\)
Chắc sai !!!
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
=>2A = 1 - 1/2187
=> A = ....
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
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