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\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=3.\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\) \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)
\(\Rightarrow2A=1-\frac{1}{3^8}\) \(\Rightarrow A=\frac{1-\frac{1}{3^8}}{2}\)
k cho mik đi mn!Nguyễn Như Quỳnh!
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}\)
\(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(3A-A=\left[1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\right]-\left[\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right]\)
\(2A=1-\frac{1}{6561}=\frac{6560}{6561}\)
\(A=\frac{6560}{6561}:2\)
\(A=\frac{3280}{6561}\)
Vậy : ...
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a) A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b) B = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
3B = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
3B - B = 1 - 1/729
2B = 728/729
B = 364/729
a, 21/11-9/19-3/13-12/38-24/22 =21/11-(9/19-3/13)-12/38-24/22 =207/143-12/38-24/22 =3075/2717-24/22 =111/2717 b,2/3+3/4+4/5+12/9+28/16+48/15 =17/12+4/5+12/9+28/16+48/15 =133/60+12/9+28/16+48/15 =71/20+28/16+48/15 =53/10+48/15 =17/2 c, 4/5-1/2-1/4-1/8-1/16-1/36 =3/10-1/4-1/8-1/36 =1/20-1/8-1/16-1/36 =-3/40-1/16-1/36 =-11/60-1/36=-119/720
Bài làm:
Ta có: \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}+\frac{1}{6561}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}+\frac{1}{3^8}\)
=> \(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}+\frac{1}{3^7}\)
=> \(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
<=> \(2A=1-\frac{1}{3^8}=\frac{3^8-1}{3^8}\)
=> \(A=\frac{3^8-1}{3^8.2}\)
Bài làm :
Ta có :
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)
\(\Rightarrow3\times A=\frac{1\times3}{3}+\frac{1\times3}{9}+\frac{1\times3}{27}+...+\frac{1\times3}{6561}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}\)
\(3\times A=1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\left(\frac{1}{6561}-\frac{1}{6561}\right)\)
\(3\times A=1+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}+\frac{1}{2187}+\frac{1}{6561}\right)-\frac{1}{6561}\)
\(3\times A=1+A-\frac{1}{6561}\)
\(\Rightarrow2\times A=1-\frac{1}{6561}\)( Trừ bỏ A ở cả 2 vế )
\(2\times A=\frac{6560}{6561}\)
\(A=\frac{6560}{6561}\div2=\frac{3280}{6561}\)
Vậy A=3280/6561
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ta gọi biểu thức trên là A
ta có :
A = \(\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{9800}\)
A = \(\frac{1}{2x4}+\frac{1}{4x6}+\frac{1}{6x8}+...+\frac{1}{98x100}\)( chữ x là dấu nhân bạn nhá ^_^"; mik ko có dấu nhân @@ )
2xA = \(\frac{2}{2x4}+\frac{2}{4x6}+\frac{2}{6x8}+...+\frac{2}{98x100}\)
=> 2 x A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
=> 2 x A = \(\frac{1}{2}-\frac{1}{100}\)=\(\frac{49}{100}\)
=> A = \(\frac{49}{100}:2\)
A = \(\frac{49}{200}\)
Ủng hộ mik nhá ^_^"
tìm số số hạng : (6561-1) / 2 +1 =3281
tổng của dãy (6561 + 1) *3281 / 2 =10764961