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a) \(5+5^2+5^3+....+5^{100}\)
đặt \(A=5+5^2+5^3+....+5^{100}\) ( \(A\) có \(100\) số hạng )
\(A=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)\) ( có \(100\div2=50\) nhóm )
\(A=5\left(1+5\right)+5^3\left(1+5\right)+....+5^{99}\left(1+5\right)\)
\(A=5.6+5^3.6+....+5^{99}.6\)
\(A=6\left(5+5^3+....+5^{99}\right)\)
vì \(6⋮6\Rightarrow6\left(5+5^3+....+5^{99}\right)⋮6\Rightarrow A⋮6\)
b) \(2+2^2+2^3+....+2^{100}\)
đặt \(B=2+2^2+2^3+....+2^{100}\) ( \(B\) có \(100\) số hạng )
\(B=\left(2+2^2+2^3+2^4+2^5\right)+.....+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\) ( có \(100\div5=20\) nhóm )
\(B=2\left(1+2+2^2+2^3+2^4\right)+....+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(B=2.31+....+2^{96}.31\)
\(B=31\left(2+...+2^{96}\right)\)
vì \(31⋮31\Rightarrow31\left(2+...+2^{96}\right)\Rightarrow B⋮31\)
a) 5+5^2+5^3..+5^100
=(5+5^2)+(5^3+5^4)+....+(5^99+5^100)
=5.(1+5)+5^3.(1+5)+....+5^99.(1+5)
=5.6+5^3.6+.....+5^99.6
=6.(5+5^3+.....+5^99):6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2+2^2\right)+...+\left(2^{99}+2^{100}\right)\)
\(A=2\cdot\left(1+2\right)+...+2^{99}\cdot\left(1+2\right)\)
\(A=2\cdot3+...+2^{99}\cdot3\)
\(A=3\cdot\left(2+...+2^{99}\right)⋮3\left(đpcm\right)\)
2 ý kia tương tự
Giải:
Đặt S=(2+2^2+2^3+...+2^100)
=2.(1+2+2^2+2^3+2^4)+2^6.(1+2+2^2+2^3+2^4)+...+(1+2+2^2+2^3+2^4).296
=2.31+26.31+...+296.31
=31.(2+26+...+296)\(⋮\)31
![](https://rs.olm.vn/images/avt/0.png?1311)
c)D=4+42+43+44+...+42012
D=(4+42)+(43+44)+...+(42011+42012)
D=4.5+43.5+45.5+...+42011.5
D=5.(4+43+42011)
=>D chia hết cho 5
=>ĐPCM