Cho S = 1 + 2 + 2^2 + 2^3 + 2^4 + 2^5 +...+ 2^2020 + 2^2021 chia hết cho
A. 10
B. 3
C. 4
D. 6
Giúp mik với
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Lời giải:
$A-1=4+4^2+4^3+...+4^{2020}+4^{2021}$
$4(A-1)=4^2+4^3+4^4+....+4^{2021}+4^{2022}$
$\Rightarrow 4(A-1)-(A-1)=4^{2022}-4$
$3(A-1)=4^{2022}-4$
$\Rightarrow 3A+1=4^{2022}\vdots 4^{2021}$
\(a,\dfrac{6}{7}+\dfrac{3}{10}=\dfrac{60}{70}+\dfrac{21}{70}=\dfrac{81}{70}\\ b,\dfrac{5}{9}+\dfrac{1}{3}=\dfrac{5}{9}+\dfrac{3}{9}=\dfrac{8}{9}\\ c,\dfrac{5}{8}-\dfrac{2}{5}=\dfrac{25}{40}-\dfrac{16}{40}=\dfrac{9}{40}\\ d,\dfrac{1}{4}-\dfrac{1}{7}=\dfrac{7}{28}-\dfrac{4}{28}=\dfrac{3}{28}\)
a) Ta có A = 1 + 21 + 22 + ... + 22021
2A = 21 + 22 + 23 + ... + 22022
Vậy 2A = 21 + 22 + 23 + ... + 22022
b) 2A - A = ( 21 + 22 + 23 + ... + 22022 ) - ( 1 + 21 + 22 + ... + 22021 )
A = 22022 - 1
Vậy A = 22022 - 1
a)
\(A=1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\)
\(2A=2^1+2^2+2^3+2^4+...+2^{2021}+2^{2022}\)
b)
\(2A=2^1+2^2+2^3+...+2^{2022}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2022}\right)-\left(1+2^1+2^2+....+2^{2021}\right)\)
\(A=2^{2022}-1\)
=> đpcm
a: \(=\dfrac{10-6}{15}\cdot\dfrac{1}{3}=\dfrac{4}{45}\)
b: \(=\dfrac{10+6}{15}\cdot\dfrac{1}{3}=\dfrac{16}{45}\)
c: \(=\dfrac{4+3}{6}\cdot\dfrac{4}{3}=\dfrac{7\cdot4}{6\cdot3}=\dfrac{28}{18}=\dfrac{14}{9}\)
d: \(=\dfrac{20-14}{35}\cdot\dfrac{9}{4}=\dfrac{6}{35}\cdot\dfrac{9}{4}=\dfrac{54}{140}=\dfrac{27}{70}\)
a (2/3 - 2/5 ) nhân 1/3
C1)
=4/15 x 1/3
= 4/45
C2)
2/3 x 1/3 - 2/5 x 1/3
= 2/9 - 2/15
= 4/45
b (2/3 + 2/5 )nhân 1/3
GIỐNG CÂU A
c (2/3 + 1/2 ) chia 3/4
C1) =7/6 : 3/4
= 14/9
C2)
2/3 : 3/4 + 1/2 : 3/4
= 8/9 + 2/3
= 14/9
d(4/7 - 2/5 ) chia 4/9
C1)
= 6/35 : 4/9
= 27/70
C2)
4/7 : 4/9 - 2/5 : 4/9
= 9/7 -9/10
= 27/70
ta có :
\(1-\frac{2}{2.3}=\frac{2.3-2}{2.3}=\frac{1.2}{2.3}\)
tương tự : \(1-\frac{2}{3.4}=\frac{2.3}{3.4},....,1-\frac{2}{2020.2021}=\frac{2019.2020}{2020.2021}\)
Vậy \(S=\frac{1.2}{2.3}.\frac{2.3}{3.4}.....\frac{2019.2020}{2020.2021}=\frac{1.\left(2.3...2019\right)^2.2020}{2.\left(3.4....2020\right)^2.2021}=\frac{2}{2020.2021}\)
Chọn B
Cảm ơn nha