Các bạn ơi , giúp mình với :
a) 1/1x3 + 1/3x5 + 1/5x7 +...+ 1/2007x2009 b) 1/18 + 1/54 + 1/108 +....+ 1/990
c) 4/2x4 + 4/4x6 + 4/4x8 +...+ 4/2008x2010 d) 32/20x23 + 32/23x26 +...+32/77x80
Ai giải nhanh nhất mình tích cho !
a) 1/1x3 + 1/3x5 + 1/5x7 +...+ 1/2007x2009
b) 3^2/20x23 + 3^2/23x26 +...+ 3^2/77x80
c) 4/2x4 + 4/4x6 + 4/4x8 +...+ 4/2008x2010
d) 1/18 + 1/54 + 1/108 +...+ 1/990
e) B= 1 + 3 +3^2 +...+ 3^100
f) A= 2^0 + 2^1 + 2^2 +...+ 2^2010
g) S= 1 + 2 + 2^2 + 2^3 +...+ 2^2008 / 1 - 2^2009
K=4/2x4+4/4x6+4/6x8+...+4/2008x2010
F=1/18+1/54+1/108+...+1/990
K = đề bài
= 2 . ( 2/2.4 + 2/4.6 + 2/6.8 + . . . + 2/2008.2010 )
= 2 . ( 1 - 1/4 + 1/4 - 1/6 + 1/8 - 1/8 + . . . + 2/2008 - 2/2010 )
= 2 . ( 1 - 2/2010 )
= ( phần còn lại bạn tự tính nha )
k cho mình đó, bài này mình làm rồi nên đúng 100% lun, sorry nha mình ngại viết nhiều
a) \(K=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(\Leftrightarrow K=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(\Leftrightarrow K=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Leftrightarrow K=2.\left(1-\frac{1}{2010}\right)\)
\(\Leftrightarrow K=2.\frac{2009}{2010}=\frac{2009}{1005}\)
b) F=1/18 + 1/54 + 1/108 +...+ 1/990
=> \(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(\Leftrightarrow F=3.\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\right)\)
\(\Leftrightarrow F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(\Leftrightarrow F=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(\Leftrightarrow F=1-\frac{1}{33}=\frac{32}{33}\)
S=1/1x3-1/2x4-1/3x5-1/4x6+1/5x7
\(S=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+\dfrac{1}{5.7}\)
\(S=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}\)
\(S=1+\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{31}{21}\)
Chúc bạn học tốt!!!
\(S=\frac{1}{1x3}-\frac{1}{2x4}+\frac{1}{3x5}-\frac{1}{4x6}+\frac{1}{5x7}-\frac{1}{6x8}+\frac{1}{7x9}-\frac{1}{8x10}\)
giúp mình với mình đang cần gấp.
\(S=\frac{1}{1.3}-\frac{1}{2.4}+\frac{1}{3.5}-\frac{1}{4.6}+\frac{1}{5.7}-\frac{1}{6.8}+\frac{1}{7.9}-\frac{1}{8.10}\)
\(\Rightarrow S=\frac{1}{2}\left(1-\frac{1}{3}-\frac{1}{2}+\frac{1}{4}+\frac{1}{3}-\frac{1}{5}-\frac{1}{4}+\frac{1}{6}+\frac{1}{5}-\frac{1}{7}-\frac{1}{6}+\frac{1}{8}+\frac{1}{7}-\frac{1}{9}-\frac{1}{8}+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}\left(1+\frac{1}{10}\right)\)
\(\Rightarrow S=\frac{1}{2}.\frac{11}{10}\)
\(\Rightarrow S=\frac{11}{20}\)
(1+1/1x3)x(1+1/2x4)x(1+1/3x5)x(1+1/4x6)x(1+1/5x7)
tìm A,B biết
A=( 1+ 1/1x3 ) x (1+1/2x4) x (1+1/3x5) x ( 1+1/4x6 ) x(1+1/5x7)
B = 1/2+1/4+1/8 +1/6 + .... + 1/128+1/256
giai duoc cai gi minh cung tick cho
Chứng minh : A = 1/1x3 + 1/2x4 + 1/3x5 + 1/4x6 + ... + 1/97x99 + 1/98x100 < 3/4
Ta có :
\(A=\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+...+\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{97}+\frac{1}{98}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-...-\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)< \frac{1}{2}.\left(1+\frac{1}{2}\right)=\frac{3}{4}\)
a, 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 +...+ 9/913 x 215
b,1/1x3 + 1/3x5 + 1/5x7 + 1/7x9 + 1/213 x 215
[ Giúp mik với mấy bạn ơi ai nhanh mình sẽ tick nha TvT ]
( Toán tính nhanh nha )
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
@ミ★Ŧɦươйǥ★彡 cảm ơn bạn nhiều
Tính tổng :
A =5+10+15+ . . . . . + 2015+2020
B = 2/1x3 + 2/3x5 + 2/5x7 + . . . . . + 2/99x101
C = 1/2x4 + 1/4x6 + 1/6x8 + . . . . . + 1/98x100
Giải nhanh giúp mk nha ! ^.^
a) Số số hạng của dãy A là: (2020-5):2+1 = 404 (số)
Tổng A là: (2020+5)x404:2=409050
b) \(B=\frac{2}{1\times3}+\frac{2}{3\times5}+....+\frac{2}{99\times101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
c) \(C=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{98\times100}\)
\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+....+\frac{2}{98\times100}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\times\left(1-\frac{1}{100}\right)=\frac{1}{2}\times\frac{99}{100}=\frac{99}{200}\)
Vậy .....
A = 5 + 10 + 15 + ... + 2015 + 2020
Số số hạng là : 404
A = 409050
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(B=1-\frac{1}{101}=\frac{101-1}{101}=\frac{100}{101}\)
\(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+...+\frac{1}{98\cdot100}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{2}\cdot\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{1}{2}\cdot\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{1}{2}\cdot\left(\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)