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\(B=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)
\(\Rightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(\Rightarrow B=\dfrac{1}{3}.\dfrac{10}{33}\)
\(\Rightarrow B=\dfrac{10}{99}\)
Vậy...
\(B=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(\Leftrightarrow B=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+..+\dfrac{1}{30.33}\)
\(\Leftrightarrow B=\left(\dfrac{1}{3}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{30}-\dfrac{1}{33}\)
\(\Leftrightarrow B=\dfrac{1}{3}-\dfrac{1}{33}\)
\(\Leftrightarrow B=\dfrac{10}{33}\).
a, A = 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/2017 - 1/2018
A = 1 - 1/2018 = 2017/2018
b, B = 5/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2016 -1/2018)
B= 5/2 . ( 1/2 - 1/ 2018 )
B = 504/1009
c, C = 1/3.6 + 1/ 6.9 + 1/ 9.12 + ... + 1/ 30.33
C= 1/3 - 1/6 + 1/6 - 1/ 9 + 1/9 - 1/12 + ... + 1/30 - 1/33
C = 1/3 - 1/33
C= 10/33
phan B mk quên nhân với 5/2
lấy 5/2 . 504/1009 = 1260/1009
a) Ta có :
\(A=1+2+2^2+2^3+....................+2^{2010}\) (\(2010\) số hạng)
\(2A=2+2^2+............+2^{2010}+2^{2011}\)
\(\Rightarrow2A-A=\left(2+2^2+..........+2^{2011}\right)-\left(1+2+.............+2^{2010}\right)\)
\(A=2^{2011}-1\)
b) Ta có :
\(B=1-3+3^2-3^3+...............+3^{100}\)(\(100\) số hạng)
\(3B=3-3^2+3^3+.....+3^{99}-3^{100}+3^{101}\)
\(\Rightarrow3B+B=\left(1-3+.......+3^{100}\right)+\left(3-3^2+....-3^{100}+3^{101}\right)\)
\(4B=3^{101}+1\)
~ Chúc bn học tốt ~
2)
\(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+\dfrac{3}{9.12}+...+\dfrac{3}{30.33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}.\dfrac{10}{33}\)
\(=\dfrac{10}{99}\)
1/18+1/54+1/108+……+1/810+1/990
=（1/3-1/6+1/6-1/9+1/9-1/12+……+1/27-1/30+1/30-1/33）÷3
=（1/3-1/33）÷3
=10/33÷3
=10/99
Ta có; F=1/3.6 +1 /6.9 + 1/9.12+......+1/30.33
F=1.3/3.6.3 + 1.3/6.9.3+......+1.3/30.33.3
F=1/3.(1/3 - 1/6 + 1/6 - 1/9 +...... +1/30 - 1/33)
F=1/3.(1/3-1/33)
F=1/3.10/33
F=10/99
giải
1/18 + 1/54 +1/108 + ......+ 1/990
ta tách mẫu số ra thành 1 tích của 2 số :
1/3x6 + 1/6x9 + 1/9x12 +........ + 1/30x33
theo quy tắc ta có : nếu tử nhân với 3 thì mẩu cũng sẽ nhân với 3 :
1x3/3x6x3 +1x3/6x9x3 + 1x3/9x11x3 + .........+ 1x3/30x33x3
= 1/3 x ( 3/3x6 + 3/6x9 + 3/9x11 +.....+3/30x33
= 1/3 x ( 1/3 - 1/33 )
= 1/3 x 10/33
=10/99
\(2.B=\dfrac{2}{6}+\dfrac{2}{14}+\dfrac{2}{60}+...+\dfrac{2}{990}\)
\(2B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{9.10.11}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{9.10}-\dfrac{1}{10.11}\)
\(2B=\dfrac{1}{1.2}-\dfrac{1}{10.11}\)
\(B=\dfrac{27}{110}\)
Ta phân tích: 1/18=1/3x6;1/54=1/6x9;1/108=1/9x12;.........1/990=1/30x33, ta có Fx3=3/3x6+3/6x9+3/9x12+........+3/30x33=1/3-1/6+1/6-1/9+1/9-1/12+.............+1/30-1/33=1/3-1/33=10/33, suy ra F là: 10/33/3=10/99
F=1/18+1/54+1/108+...+1/990 F=1/3.6 + 1/6.9 + 1/9.12 +...+ 1/30.33 suy ra : 3F= 3/3.6 + 3/6.9 + 3/9.12 +...+3/30.33 3F= 3/3 - 3/6 + 3/6 - 3/9 + 3/9 - 3/12 +...+3/30 - 3/33 3F=1 - 3/33 = 33/33 - 3/33 = 30/33 F= 30/33 : 3 = 30/33 . 1/3 =10/99
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{10}{33}\)
\(F=\frac{10}{99}\)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
Thử làm xem , không biết đúng không nhé
!
\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\)
\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(F=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\)
\(F=\dfrac{1}{3}.\dfrac{10}{33}\)
\(F=\dfrac{10}{99}\)
\(F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+.....+\dfrac{1}{990}\)
\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+......+\dfrac{1}{30.33}\)
\(F=\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+.....+\dfrac{1}{30}-\dfrac{1}{33}\)
\(F=\dfrac{1}{3}-\dfrac{1}{33}\)
\(F=\dfrac{10}{33}\)