đặt \(A=2+2^2+2^3+....+2^{100}\)
+) \(A=2+2^2+2^3+.....+2^{100}\)
\(A\) có : \(\left(100-1\right)\div1+1=100\) ( số hạng )
\(A=\left(2+2^2+2^3+2^4+2^5\right)+....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\) ( \(A\) có \(100\div5=20\)nhóm )
\(A=2\left(1+2+2^2+2^3+2^4\right)+.....+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(A=2.31+.....+2^{96}.31\)
\(A=31\left(2+.....+2^{96}\right)\)
vì \(31⋮31\Rightarrow31\left(2+....+2^{96}\right)⋮31\)
\(\Rightarrow A⋮31\)
+) \(A=2+2^2+2^3+.....+2^{100}\)
\(A=\left(2+2^2+2^3+2^4\right)+....+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\) ( \(A\) có : \(100\div4=25\) nhóm )
\(A=2\left(1+2+2^2+2^3\right)+.....+2^{97}\left(1+2+2^2+2^3\right)\)
\(A=2.15+.....+2^{97}.15\)
\(A=15\left(2+....+2^{97}\right)\)
vì \(15⋮5\Rightarrow15\left(2+....+2^{97}\right)⋮5\)
\(\Rightarrow A⋮5\)
C = 2 + 2^2 + 2^3 + …….. + 2^99 + 2^100
= 2(1 +2 + 2^2+ 2^3+ 2^4) + 2^6(1 + 2 + 2^2+ 2^3+ 2^4)+…+ (1 + 2 + 22+ 23+ 24).2^96
= 2 . 31 + 2^6 . 31 + … + 2^96 . 31
= 31(2 + 2^6 +…+29^6) \(⋮\)31
Vậy C chia hết cho 31.
(2+2^2+2^3+2^4)+....+(2^97+2^98+2^99+2^100)
=> A= 1.(2+2^2+2^3+2^4)+.............+2^96.(2+22+23+24)
=> A= 1.30+........+2^96.30
=> A= 30.(1+.......+2^96) chia hết cho 5
Vậy A \(⋮\)5
Vậy 2+2^2+2^3+...+2^100 vừa chia hết cho 31 ,vừa chia hết cho 5
Chia hết cho 31 là sai rồi bn xem lại đề đi bn
2+2^2+2^3+2^4+2^5+...+2^100 chia hết cho 5
(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+...+(2^97+2^98+2^99+2^100)
2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+..+2^97(1+2+2^2+2^3)
2.15+2^5.15+..+2^97.15
15.(2+2^5+..+2^97)
3.5(2+2^5+...+2^97)
<=> đcpm
