\(S=\frac{1}{5\times6}+\frac{1}{10\times9}+\frac{1}{15\times12}+...+\frac{1}{3350\times2013}\)
K
Khách
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Những câu hỏi liên quan
21 tháng 10 2016
Ta có: \(\frac{1}{4\times6}=\frac{1}{4\times1\times3\times2}=\frac{1}{4\times3\times1\times2}\)
\(\frac{1}{8\times9}=\frac{1}{4\times2\times3\times3}=\frac{1}{4\times3\times2\times3}\)
\(\frac{1}{12\times12}=\frac{1}{4\times3\times3\times4}\)
\(\frac{1}{16\times15}=\frac{1}{4\times4\times3\times5}=\frac{1}{4\times3\times4\times5}\)......
\(\frac{1}{2680\times2013}=\frac{1}{4\times670\times3\times671}\)
Do đó:
\(M=\frac{1}{4\times3}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+....+\frac{1}{670\times671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{670}-\frac{1}{671}\right)\)
\(=\frac{1}{12}\times\left(\frac{1}{1}-\frac{1}{671}\right)=\frac{1}{12}\times\frac{670}{671}=\frac{335}{4026}\)
Vậy \(M=\frac{335}{4026}\)
TQ
26 tháng 3 2020
Ta có :
1/2x6 + 1/4x9 + 1/6x12 +...+1/198 x 300
= 1/6x2 + 1/6x6 + 1/6x12 + ....+1/6x9900
= 1/6 x ( 1/2 + 1/6 + 1/ 12 +...+1/9900)
= 1/6 x (1/1x2 + 1/2x3 + 1/3x4+...+1/99x100)
=1/6x (1-1/2 + 1/2-1/3 + 1/3 - 1/4 + ....+1/99-1/100)
=1/6x(1-1/100)
=1/6 x 99/100
= 33/200
k cho mình nha , học tốt
6 tháng 8 2016
\(\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{38.60}\)
\(=\frac{1}{2.3}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{6}.\frac{19}{20}=\frac{19}{120}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
6 tháng 6 2023
A = \(\dfrac{1}{3\times6}\) + \(\dfrac{1}{6\times9}\) + \(\dfrac{1}{9\times12}\)+...+\(\dfrac{1}{144\times147}\)
A = \(\dfrac{1}{3}\) \(\times\)( \(\dfrac{3}{3\times6}\) + \(\dfrac{3}{6\times9}\)+\(\dfrac{1}{9\times12}\)+...+\(\dfrac{3}{144\times147}\))
A = \(\dfrac{1}{3}\) \(\times\)(\(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{144}-\dfrac{1}{147}\))
A = \(\dfrac{1}{3}\)\(\times\)(\(\dfrac{1}{3}\) - \(\dfrac{1}{147}\))
A = \(\dfrac{1}{3}\) \(\times\)\(\dfrac{16}{49}\)
A = \(\dfrac{16}{147}\)
24 tháng 5 2019
\(A=\frac{1}{1\times6\times6}+\frac{1}{2\times9\times8}+\frac{1}{3\times12\times10}+...+\frac{1}{98\times297\times200}\)
\(A=\frac{1}{1\times\left(2\times3\right)\times\left(2\times3\right)}+\frac{1}{2\times\left(3\times3\right)\times\left(2\times4\right)}+...+\frac{1}{98\times\left(99\times3\right)\times\left(100\times2\right)}\)
\(A=\frac{1}{6}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{98\times99\times100}\right)\)
\(12\times A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{98\times99\times100}\)
\(12\times A=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{98\times99}-\frac{1}{99\times100}\right)\)
\(12\times A=\frac{1}{1\times2}-\frac{1}{99\times100}=\frac{4949}{9900}\)
\(A=\frac{4949}{118800}\)
HD
4
5 tháng 8 2016
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{5}{15}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
Dấu . là nhân nha
5 tháng 8 2016
\(\frac{4}{3.6}+\frac{4}{6.9}+\frac{4}{9.12}+\frac{4}{12.15}\)
\(=\frac{4}{3}.\left(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+\frac{3}{12.15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+\frac{1}{12}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{3}-\frac{1}{15}\right)\)
\(=\frac{4}{3}.\frac{4}{15}=\frac{16}{45}\)
S = \(\frac{1}{5.6}+\frac{1}{10.9}+\frac{1}{15.12}+...+\frac{1}{3350.2013}=\frac{1}{5.3}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{670.671}\right)\)
\(=\frac{1}{15}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{670}-\frac{1}{671}\right)=\frac{1}{15}\left(1-\frac{1}{671}\right)=\frac{1}{15}.\frac{670}{671}\)
\(=\frac{134}{2013}\)