DD
0
K
Khách
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DD
1
22 tháng 3 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{19}+\frac{1}{20}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
\(=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\)
LT
0
LT
0
AK
12 tháng 3 2018
1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/19 - 1/20
= ( 1 + 1/3 + 1/5 + ...+ 1/19 ) - ( 1/2 + 1/4 + ...+ 1/20 )
= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/19 + 1/20 ) - 2 . ( 1/2 + 1/4 + ...+ 1/20 )
= ( 1 + 1/2 + 1/3 + ...+ 1/20 ) - ( 1 + 1/2 + ... + 1/10 )
= 1/11 + 1/12 + 1/13 + ...+ 1/20 ( Đpcm )
TK mk nha !!!
12 tháng 3 2018
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{19}-\frac{1}{20}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{19}+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}-1+\frac{1}{2}+....+\frac{1}{10}\)
\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\left(đpcm\right)\)
NM
0
GB
Gia Bảo
VIP
1 tháng 3
M=1−31+1−151+1−351+1−631+...+1−99991
\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + . . . + \frac{1}{9999} \left.\right)\)
\(� = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + . . . + \frac{1}{99.101} \left.\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)
\(� = 50 - \frac{1}{2} . \left(\right. \frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \frac{2}{7.9} + . . . + \frac{2}{99.101} \left.\right)\)
\(� = 50 - \left(\right. 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + . . . + \frac{1}{99} - \frac{1}{101} \left.\right)\)
\(� = 50 - \left(\right. 1 - \frac{1}{101} \left.\right) = 50 - \frac{100}{101} = \frac{5050 - 100}{101} = \frac{4950}{101}\)
BT
22 tháng 3 2018
Ta có: \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)=\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)=\)
= 1/11 + 1/12 +1/13+...+1/20 (đpcm)