NM
1
K
Khách
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17 tháng 4 2016
D = -1-1/3-1/6-1/10-...-1/1225
Suy ra : D/2=-1/2-1/6-1/12-....-1/2450
Mà 1/2=1/(1.2)=1-1/2; 1/6=1/(2.3)=1/2-1/3;...1/2450=1/(49.50)=...
D/2= -(1-1/2)-(-1/2-1/3)-...-(1/49-1/50)
D/2= -1+1/2-1/2+1/3-....-1/49+1/50
D/2= -1+1/50=-49/50
D=(-49/50).2=-98/50
k nha
TN
1
DT
2
VG
12 tháng 12 2023
A = -1-\(\dfrac{1}{3}\)-\(\dfrac{1}{6}\)-\(\dfrac{1}{10}\)-\(\dfrac{1}{15}\)-...-\(\dfrac{1}{1225}\)
= -1-(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\))
Đặt B = \(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+...+\(\dfrac{1}{1225}\)
Ta có : B = 2(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{2450}\))
= 2(\(\dfrac{1}{2\text{×}3}\)+\(\dfrac{1}{3\text{×}4}\)+\(\dfrac{1}{4\text{×}5}\)+\(\dfrac{1}{5\text{×}6}\)+...+\(\dfrac{1}{49\text{×}50}\))
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{49}\)-\(\dfrac{1}{50}\)
= 2(\(\dfrac{1}{2}\)-\(\dfrac{1}{50}\))
= 2×\(\dfrac{24}{50}\)
= \(\dfrac{24}{25}\)
Thay B vào A ta có :
A = -1-\(\dfrac{24}{25}\)
=> A = \(\dfrac{-49}{25}\)
Cho mik một tick nhé thankss
HT
2
14 tháng 9 2023
\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)
\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)
\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)
\(\dfrac{1}{2}B=\dfrac{-49}{50}\)
\(B=\dfrac{-49}{25}\)
14 tháng 9 2023
\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)
\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=-2*49/50
=-49/25
30 tháng 6 2022
\(=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=-2\cdot\dfrac{49}{50}=\dfrac{-49}{25}\)
LN
2
S
10 tháng 7 2018
\(B=-1-\frac{1}{3}-\frac{1}{6}-...-\frac{1}{1225}\)
\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)
10 tháng 7 2018
\(B=-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
\(B=-2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{49\cdot50}\right)\)
\(B=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(B=-2\left(1-\frac{1}{50}\right)\)
\(B=-2\cdot\frac{49}{50}\)
\(B=-\frac{49}{25}\)
NP
1
20 tháng 7 2017
\(-1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-....-\frac{1}{1225}\)
\(=-2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{2450}\right)\)
\(=-2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\right)\)
\(=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=-2\left(1-\frac{1}{50}\right)=-2\cdot\frac{49}{50}=-\frac{49}{25}\)
DD
1
4 tháng 7 2017
\(-1-\dfrac{1}{3}-\dfrac{1}{6}-...-\dfrac{1}{1225}\)
\(=\dfrac{-1}{2}\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2450}\right)\)
\(=\dfrac{-1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{-1}{2}\left(1-\dfrac{1}{50}\right)\)
\(=\dfrac{-1}{2}.\dfrac{49}{50}=\dfrac{-49}{100}\)
Vậy...
20 tháng 6 2019
Đặt : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1225}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{2450}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{49.50}\)
\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(A=2.\frac{12}{25}\)
\(A=\frac{24}{25}\)
\(F=-1-\dfrac{1}{3}-\dfrac{1}{6}-...-\dfrac{1}{1225}\)
\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=-2\left(1-\dfrac{1}{50}\right)=-2\cdot\dfrac{49}{50}=-\dfrac{49}{25}\)