½ +1/22 + 1/23 + … + ½20 <1
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DM
1
23 tháng 3 2023
1/20 .21 + 1/22 .23 + .... + 1/79 .80
= 1/20 - 1/21 + 1/22 - 1/23 + .......... + 1/79 - 1/80
= 1/20 - 1/80
= 3/80
Ta thấy : 3/80 < 1
=> 1/20 . 21 + 1/22 . 23 + ........ + 1/79 . 80 <1 (ĐPCM)
14 tháng 12 2022
a: \(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{48}\left(1+2\right)\)
\(=3\left(1+2^2+...+2^{48}\right)⋮3\)
b: \(2^0+2^1+2^2+...+2^{101}\)
\(=\left(1+2+2^2\right)+...+2^{99}\left(1+2+2^2\right)\)
\(=7\left(1+...+2^{99}\right)⋮7\)
c: 2A=2+2^2+...+2^101
=>A=2^101-1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
5 tháng 4 2023
a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)
= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)
= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)
= 1 - 1 + \(\dfrac{2022}{2023}\)
= \(\dfrac{2022}{2023}\)
b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)
= \(\dfrac{33}{11}\)
= 3
c, 2000 + { 20 - [ 4.20220 - (32 + 5):2] }
= 2000 + { 20 - [ 4.1 - (9+5):2]}
= 2000 + { 20 - [ 4 - 14 : 2 ]}
= 2000 + { 20 - [ 4 -7]}
= 2000 + { 20 - (-3)}
= 2000 + 23
= 2023
NT
0
LH
2
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
\(\Rightarrow2A-A=(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}})-(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{20}})\)
\(\Rightarrow A=1-\frac{1}{2^{20}}< 1\)