cho P=1/3+1/6+1/10+...1/105
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TT
1
MA
30 tháng 3 2016
nhân 2 vế với 1/2 ta có
1/2 x A = 1/2 x (1/3 + 1/6 +1/10 + 1/15 + .......+1/91 + 1/105 )
1/2 x A = 1/6 +1/12 + 1/20 + 1/30 + ...............+1/182 + 1/210
1/2 x A = 1/(2x3) + 1/(3x4) + 1/(4x5) + 1/(5x6) +................+1/(13x14) + 1/(14x15)
1/2 x A = 1/2 - 1/3 +1/3 -1/4 + 1/4 - 1/5 +1/5 - 1/6+.........+1/13 - 1/14 + 1/14 - 1/15
1/2 x A = 1/2 - 1/15 =13/30
=> A = 13/30 : 1/2=13/15 <1
HT
24 tháng 7 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{105}\)
=\(\frac{2}{6}+\frac{2}{12}+\frac{1}{20}+.....+\frac{2}{210}\)
= \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{14.15}\right)\)
= \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{14}-\frac{1}{15}\right)\)
= \(2\left(\frac{1}{2}-\frac{1}{15}\right)\)
= 2 . \(\frac{13}{30}\)
= \(\frac{13}{15}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 4 2023
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)+.....+ \(\dfrac{1}{105}\)
A \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) +.....+ \(\dfrac{1}{105}\))
A \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+.....+ \(\dfrac{1}{210}\)
A \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+ \(\dfrac{1}{4\times5}\)+....+\(\dfrac{1}{14\times15}\)
A \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{1}{2}-\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+.....+ \(\dfrac{1}{14}\) - \(\dfrac{1}{15}\)
A \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{15}\)
A \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{13}{30}\)
A = \(\dfrac{13}{30}\) : \(\dfrac{1}{2}\)
A = \(\dfrac{13}{15}\)
KT
9 tháng 8 2018
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{105}+\frac{1}{210}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{210}+\frac{1}{240}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{14.15}+\frac{1}{15.16}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{!}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)
\(=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
=> \(A=\frac{7}{8}\)
AH
0
TP
2
27 tháng 4 2017
\(M=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{105}+\frac{1}{120}\)
\(M=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{210}+\frac{2}{240}\)
\(M=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{14.15}+\frac{2}{15.16}\)
\(M=\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+\frac{2}{6}-\frac{2}{7}+\frac{2}{7}-\frac{2}{8}+...+\frac{2}{15}-\frac{2}{16}\)
\(M=\frac{2}{4}-\frac{2}{16}=\frac{3}{8}\)
Vì \(\frac{3}{9}< \frac{3}{8}< \frac{4}{8}\)nên \(\frac{1}{3}< M< \frac{1}{2}\)
Vậy \(\frac{1}{3}< M< \frac{1}{2}\)
P/S : Đừng nói như lần trước nhé!
CM
20 tháng 9 2018
C=3-1+4-1+5-1+....+102-1+103-1
C=2+3+4+5+...+101+102
Tổng đã cho có: (102-2):1+1=101 (số hạng)
C=(102+2)*101:2=5252
Vậy 5252 là tổng của tập hợp C.
20 tháng 9 2018
C = 3 - 1 + 4 - 1 + 5 - 1 + .... + 102 - 1 + 103 - 1
= 2 + 3 + 4 + ... + 101 + 102
Số số hạng là : (102 - 2) : 1 + 1 = 100 (số hạng)
Tổng : (102 + 2) . 100 : 2 = 5200
Vậy C = 5200
P = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)+.....+ \(\dfrac{1}{105}\)
P = \(\dfrac{2}{2}\) \(\times\) ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)+ ......+ \(\dfrac{1}{105}\))
P = 2 \(\times\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+......+ \(\dfrac{1}{210}\))
P = 2 \(\times\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + \(\dfrac{1}{4\times5}\)+.....+\(\dfrac{1}{14\times15}\))
P = 2 \(\times\) ( \(\dfrac{1}{2}-\dfrac{1}{3}\) + \(\dfrac{1}{3}-\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +......+ \(\dfrac{1}{14}\) - \(\dfrac{1}{15}\))
P = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{15}\))
P = 2 \(\times\) \(\dfrac{13}{30}\)
P = \(\dfrac{13}{15}\)