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Đặt \(S=1+2^1+2^2+2^3+2^4+...+2^{20}\)
\(=2^0+2^1+2^2+2^3+2^4+2^{19}\) ( tong cap so nhan co 20 so hang. cong boi q=2.u1=1)
\(\Rightarrow s=\frac{u1.\left(1-q^{20}\right)}{\left(1-q\right)}=\frac{\left(1-2^{20}\right)}{\left(1-2\right)}=10485...\)
A=1+2^1+2^2+...+2^20
=>2A=2+2^2+2^3+.....+2^20+2^21
=>2A - A=(2+2^2+...+2^21)-(1+2+2^2+...+2^20)
hay A=2^21-1
Đặt \(A=1+2+2^2+2^3+...+2^{20}\)
\(2A=2+2^2+2^3+2^4+...+2^{21}\)
\(2A-A=\left(2+2^2+2^3+...+2^{21}\right)-\left(1+2+2^2+...+2^{20}\right)\)
\(A=2^{21}-1\)
Ta đặt
A= 1+2^1+2^2+2^3+....2^20
2A= 21+22+23+....+221
=>2A-A=(2^1+2^2+2^3+...+2^21)-(1+2^2+2^3+...)
1A=2^21-1
Vậy A=2^21-1
1+2^1+2^2+2^3+2^4+...+2^20
=2*(1+2^1+2^2+2^3+2^4+...+2^20)-1-2^1+2^2-2^3-2^4-...-2^20
=2^1+2^2+2^3+2^4+...+2^21-1-21-2^2-2^3-2^4-...-2^21
=2^21-1
S= (5 +5^2+5^3) +(5^4+5^5+5^6)+...+(5^2017+5^2018+5^2019)
=5(1+5+5^2)+5^4(1+5+5^2)+...+5^2017(1+5+5^2)
=5.31+5^4.31+...+5^2017.31
=31.( 5+5^4+...+5^2017) chia hết cho 31 (đpcm)
\(S=5+5^1+5^2+5^3+...+5^{2024}\)
\(=5+\left(5^1+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2021}+5^{2022}+5^{2023}+5^{2024}\right)\)
\(=5+\left(5^1+5^2+5^3+5^4\right)+5^4\left(5^1+5^2+5^3+5^4\right)+...+5^{2020}\left(5^1+5^2+5^3+5^4\right)\)
\(=5+780\left(1+5^4+...+5^{2020}\right)\)
Có \(780⋮65\)nên \(780\left(1+5^4+...+5^{2020}\right)⋮65\)
suy ra \(S\)chia cho \(65\)dư \(5\).
22 . 31(12018.20180) : |-2|
= 4 .3(1 . 1) : 2
=6
|-8| - [42 + (-5)] +(-17)
= 8 -[16+ (-5)] + (-17)
=8 - 11 + (-17)
=-20
chúc hok tốt
a . 4.3 - (1+1) : 2
=12 - 2 : 2
= 12-1
= 11
b 8 - [16+(-5)]+(-17)
=8-11+(-17)
=(-3)+(-17)
= -20
20180 = 1
52 = 25
Học tốt
Nhớ t.i.c.k
#Vii
20180 = 1
52 = 25
K cho mk nhé!