chung minh S= 2+22+23+...+2100 chia het cho 15
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1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{97}\right)\)
\(=30\left(1+2^4+...+2^{96}\right)⋮30\)
2:
\(B=3+3^2+3^3+...+3^{2022}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=(2+2^2)+(2^3+2^4)+....+(2^{99}+2^{100})$
$=2(1+2)+2^3(1+2)+...+2^{99}(1+2)$
$=2.3+2^3.3+...+2^{99}.3$
$=3(2+2^3+...+2^{99})\vdots 3$
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A+2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+...+2^{98}\left(2+2^2\right)\)
\(=6+2^2.6+...+2^{98}.6=6\left(1+2^2+...+2^{98}\right)⋮6\)
\(A=2+2^2+2^3+2^4+...+2^{100}\)
\(=2\cdot3+...+2^{99}\cdot3\)
\(=6\left(1+...+2^{99}\right)⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có : \(S=1+2+2^2+2^3+....+2^{99}\)
\(\Rightarrow2S=2+2^2+2^3+....+2^{100}\)
\(\Rightarrow2S-S=\left(2+2^2+2^3+...+2^{100}\right)-\left(1+2+2^2+....+2^{99}\right)\)
\(\Rightarrow S=2^{100}-1< 2^{100}\)
Vậy \(S< 2^{100}\)
S=1+2+22+23+....+299
⇒2S=2+22+23+....+2100
⇒2S−S=2100-1
S=2100-1
vì 2100 -1<2100
⇒S<2100
![](https://rs.olm.vn/images/avt/0.png?1311)
*Sửa lại đề*
A = 21+ 22+ 23+ 24 + .. + 2100
A = (21+22) + (23+ 24) +...+ (299+ 2100)
A = 2.(1+2) + 23.(1+2) + .. + 299. (1+2)
A = 2.3 + 23. 3 + .. + 299.3
A = 3 . (21 + 23 + .... + 299)
Mà 3 chia hết cho 3
=> A chia hết cho 3
Lời giải:
Ta có:
$S=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+....+(2^{97}+2^{98}+2^{99}+2^{100})$
$=2(1+2+2^2+2^3)+2^5(1+2+2^2+2^3)+....+2^{97}(1+2+2^2+2^3)$
$=(1+2+2^2+2^3)(2+2^5+...+2^{97})$
$=15(2+2^5+....+2^{97})\vdots 15$