Tính tổng S = 10.(\(\frac{1}{2.12}\)+\(\frac{1}{12.22}\)+\(\frac{1}{22.32}\)+\(\frac{1}{32.42}\)+.........+\(\frac{1}{2002.2012}\))
S= (1/2.12+1/12.22+.....1/2002.2012)
các bạn ơi giải giùm mk nhá
\(S=\frac{1}{2\cdot12}+\frac{1}{12\cdot22}+...+\frac{1}{2002\cdot2012}\)
\(S=\left(\frac{1}{2\cdot12}+\frac{1}{12\cdot22}...+\frac{1}{2002\cdot2012}\right)\cdot\frac{1}{10}\)
\(S=\left(\frac{1}{2}-\frac{1}{12}+\frac{1}{12}-\frac{1}{22}+...+\frac{1}{2002}-\frac{1}{2012}\right)\cdot\frac{1}{10}\)
\(S=\left(\frac{1}{2}-\frac{1}{2012}\right)\cdot\frac{1}{10}\)
\(S=\frac{1006-1}{2012}\cdot\frac{1}{10}\)
\(S=\frac{1005}{2012}\cdot\frac{1}{10}\)
\(S=\frac{201}{2012}\cdot\frac{1}{2}\)
\(S=\frac{201}{4024}\)
S=1/10 \(\times\)( 1/2 - 1/12 + 1/12 - 1/22 +...+1/2002 -1/2012 )
= 1/10 \(\times\)( 1/2 - 1/2012)
= 1/10 \(\times\)1005/2012
= 201/4024
Chứng minh rằng: a, 1/12.22+5/22.32+5/32.42+...+5/92.102 <1 b,1/3+2/32+3/33+...+100/3100 <3/4
Đây Là Lớp Mấy
Bài 1 : Tính tổng S , biết : \(S=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{2010\times2011}\)
Bài 2 : Tính tổng sau : \(S=\frac{3}{10\times13}+\frac{3}{13\times16}+\frac{3}{16\times19}+....+\frac{3}{58\times61}\)
Bài 3 : Tính tổng sau : \(S=\frac{1}{4\times7}+\frac{1}{7\times10}+\frac{1}{10\times13}+....+\frac{1}{19\times22}\)
Bài 1 :
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\)
\(S=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
Bài 2 :
\(S=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)
\(S=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)
Bài 3 :
\(3S=\frac{3}{4\times7}+\frac{3}{7\times11}+...+\frac{3}{19\times22}\)
\(3S=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{19}-\frac{1}{22}\)
\(3S=\frac{1}{4}-\frac{1}{22}\)
\(S=\frac{18}{88}\div3=\frac{6}{88}\)
Giúp mình với các bạn::
Chứng minh:
a) \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{10}\left(\frac{1}{n}-\frac{1}{n+10}\right)=\frac{1}{n\left(n+10\right)}\)
c)Tổng quát: \(\frac{1}{a}\left(\frac{1}{n}-\frac{1}{n+a}\right)=\frac{1}{n\left(n+a\right)}\)
Áp dụng tính tổng sau: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\)
Áp dụng giải phương trình sau:
\(\left(\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{10.110}\right)x=\left(\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{100.10}\right)\)
cau a),b),c) ban dat mau chung roi khu mau ma lam la duoc ma
Tuy học lớp 6 ................. cơ mừ thấy mí bài nỳ dễ quá >.<
Tính:
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+\frac{1}{2.3}+\frac{5}{12.17}+\frac{6}{17.23}+\frac{7}{23.30}\)
đề sai thì phải
\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)
\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)
\(A=\frac{101}{120}\)
\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)
\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)
\(E=\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+.........+\frac{1}{10.110}\)
\(F=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+.........+\frac{1}{100.110}\)
tính tỉ số \(\frac{E}{F}\)
\(100E\)\(=\frac{100}{1.101}+\frac{100}{2.102}+..........+\frac{100}{10.110}\)
\(=1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+........+\frac{1}{10}-\frac{1}{110}\)
\(10F=\frac{10}{1.11}+\frac{10}{2.12}+......+\frac{10}{100.110}\)
\(=1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+......+\frac{1}{100}-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}+\frac{1}{11}+....+\frac{1}{100}-\frac{1}{11}-\frac{1}{12}-....-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{110}\)
\(=1+\frac{1}{2}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\)\(=100E\)
\(\Rightarrow10F=100E\Rightarrow\frac{E}{F}=\frac{1}{10}\)
cho : \(A=\frac{1}{1.51}+\frac{1}{2.52}+\frac{1}{3.53}+...+\frac{1}{10.60}\)và \(B=\frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{50.60}\)
tính \(\frac{B}{A}\)
Tính tổng S 10. \ \frac{1}{1.11}\ \ \frac{1}{11.21}\ \ \frac{1}{21.31}\ \ \frac{1}{31.41}\ ... \ \frac{1}{91.101}\
Mong anh chị em giúp em với ạ
Tính tỉ số A/B biết:
A\(=\frac{1}{1.101}+\frac{1}{2.102}+...\frac{1}{10.110}\)
B\(=\frac{1}{1.11}+\frac{1}{2.12}+...\)