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=1/1x2-1/2x3+1/2x3-1/3x4+...+1/98x99-1/99x100
=1/2-1/9900
=4949/9900
Đặt C = \(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+...+\frac{1}{30\times31\times32}\)
\(2C=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+...+\frac{2}{30\times31\times32}\)
\(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)
\(=\frac{1}{1\times2}-\frac{1}{31\times32}\)
\(=\frac{1}{2}-\frac{1}{992}=\frac{495}{992}\)
\(\Rightarrow C=\frac{495}{992}\div2=\frac{495}{1984}\)
Vậy ...
\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+.....+\frac{1}{30\times31\times32}\)
\(=\frac{1}{2}\times\left(\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+.....+\frac{2}{30\times31\times32}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{30.31}-\frac{1}{31.32}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{31.32}\right)=\frac{1}{2}.\frac{990}{1984}=\frac{990}{3968}\)
Đặt \(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{30\times31\times32}\)
\(=\left(\frac{1}{1\times2}-\frac{1}{2\times3}\right)+\left(\frac{1}{2\times3}-\frac{1}{3\times4}\right)+\left(\frac{1}{3\times4}-\frac{1}{4\times5}\right)+...+\left(\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\)
\(=\frac{1}{1\times2}-\frac{1}{31\times32}\)
\(=\frac{1}{2}-\frac{1}{992}\)
\(=\frac{495}{992}\)
\(\Rightarrow A=\frac{495}{992}\div2=\frac{495}{1984}\)
\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{30\times31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{30\times31}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{31\times32}\right)\)
\(=\frac{1}{2}\times\frac{990}{1984}\)
\(=\frac{990}{3968}=\frac{495}{1984}\)
Sửa đề chút:
\(\frac{1}{1x2x3}+\frac{1}{2x3x4}+...+\frac{1}{98x99x100}\)
\(=\frac{1}{2}.\left(\frac{2}{1x2x3}+\frac{2}{2x3x4}+...+\frac{2}{98x99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1x2}-\frac{1}{2x3}+\frac{1}{2x3}-\frac{1}{3x4}+...+\frac{1}{98x99}-\frac{1}{99x100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{4}-\frac{1}{99.200}< 1\)
đpcm
A = 3/1.3 + 3/3.5 + 3/5.7 + 3/7.9 + ... + 3/97.99
A = 3/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + .... + 2/97 - 2/99
A = 3/2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 )
A = 3/2 . ( 1 - 1/99 )
A = 3/2 . 98/99
A = 49/33
b) dãy số không có quy luật==> bạn xem lại đề
c) \(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{50\times51}-\frac{1}{51\times52}\right)\)
\(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{51\times52}\right)=\frac{1}{2}\times\frac{2650}{5408}=\frac{1325}{5408}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{50.51}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2550}\right)\)
\(=\frac{1}{2}.\left(\frac{1275}{2550}-\frac{1}{2550}\right)\)
\(=\frac{1}{2}.\frac{1274}{2550}\)
\(=\frac{637}{2550}\)
Lưu ý : Dấu \("."\)là dấu \("\)x \("\)
( dấu nhân )
Chúc bạn học giỏi !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
VD ( dễ hiểu )
A = \(\dfrac{2}{1\times3\times5}\) + \(\dfrac{2}{3\times5\times7}\) + \(\dfrac{2}{5\times7\times9}\)+\(\dfrac{2}{7\times9\times11}\)
A = \(\dfrac{1}{2}\) x (\(\dfrac{4}{1\times3\times5}\) + \(\dfrac{4}{3\times5\times7}\) + \(\dfrac{4}{5\times7\times9}\) + \(\dfrac{4}{7\times9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\)-\(\dfrac{1}{3\times5}\)+\(\dfrac{1}{3\times5}\)-\(\dfrac{1}{5\times7}\)+\(\dfrac{1}{5\times7}\)-\(\dfrac{1}{7\times9}\)+\(\dfrac{1}{7\times9}\)-\(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\)x (\(\dfrac{1}{1\times3}\) - \(\dfrac{1}{9\times11}\))
A = \(\dfrac{1}{2}\) x (\(\dfrac{1}{3}-\dfrac{1}{99}\))
A = \(\dfrac{1}{2}\times\) \(\dfrac{32}{99}\)
A = \(\dfrac{16}{99}\)
B = \(\dfrac{1}{1\times2\times3}\) + \(\dfrac{1}{2\times3\times4}\) + \(\dfrac{1}{3\times4\times5}\) + \(\dfrac{1}{4\times5\times6}\)
B = \(\dfrac{1}{2}\) x (\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}\))
B = \(\dfrac{1}{2}\) x (\(\dfrac{1}{1\times2}\)-\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times3}\)-\(\dfrac{1}{3\times4}\)+\(\dfrac{1}{3\times4}\)-\(\dfrac{1}{4\times5}\)+\(\dfrac{1}{4\times5}\)-\(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x(\(\dfrac{1}{1\times2}\) - \(\dfrac{1}{5\times6}\))
B = \(\dfrac{1}{2}\)x (\(\dfrac{1}{2}-\dfrac{1}{30}\))
B = \(\dfrac{1}{2}\)x \(\dfrac{7}{15}\)
B = \(\dfrac{7}{30}\)
\(A=\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{36\times37\times38}+\frac{1}{37\times38\times39}\)
\(2A=\frac{2}{1\times2\times3}+\frac{2}{2\times3\times4}+\frac{2}{3\times4\times5}+...+\frac{2}{36\times37\times38}+\frac{2}{37\times38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{37\times38}-\frac{1}{38\times39}\)
\(2A=\frac{1}{1\times2}-\frac{1}{38\times39}\)
\(2A=\frac{741}{1482}-\frac{1}{1482}\)
\(2A=\frac{370}{741}\)
\(A=\frac{370}{741}:2=\frac{185}{741}\)
`A=1/[1xx2xx3]+1/[2xx3xx4]+1/[3xx4xx5]+....+1/[98xx99xx100]`
`A=1/2xx(2/[1xx2xx3]+2/[2xx3xx4]+2/[3xx4xx5]+....+2/[98xx99xx100])`
`A=1/2xx(1/[1xx2]-1/[2xx3]+1/[2xx3]-1/[3xx4]+1/[3xx4]-1/[4xx5]+....+1/[98xx99]-1/[99xx100])`
`A=1/2xx(1/[1xx2]-1/[99xx100])`
`A=1/2xx(1/2-1/9900)`
`A=1/2xx(4950/9900-1/9900)`
`A=1/2xx4949/9900`
`A=4949/19800`
\(A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}\)
\(A=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right):2\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{12}-\dfrac{1}{20}+...+\dfrac{1}{9702}-\dfrac{1}{990}\right):2\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{990}\right):2\)
\(A=\dfrac{4949}{9900}:2\)
\(A=\dfrac{4949}{19800}\)