Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+5+5^2+5^3+...+5^{59}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{57}+5^{58}+5^{59}\right)\)
\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+...+5^{57}\left(1+5+5^2\right)\)
\(=31\left(1+5^3+...+5^{57}\right)\)chia hết cho \(31\).
\(A=1+5+5^2+5^3+...+5^{59}\)
\(5A=5+5^2+5^3+5^4+...+5^{60}\)
\(5A-A=\left(5+5^2+5^3+5^4+...+5^{60}\right)-\left(1+5+5^2+5^3+...+5^{59}\right)\)
\(4A=5^{60}-1\)
\(A=\frac{5^{60}-1}{4}< \frac{5^{60}}{4}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
S=1+51+52+53+...+559
S=(1+51+52)+(53+54+55)+....+(557+558+559)
S=31+53.(1+5+52)+....+557.(1+5+52)
S=31+53.31+...+557.31
S=31.(1+53+...+557)
vì 31 chia hết cho 31 nên S chia hết cho 31
vậy S chia hết cho 31
![](https://rs.olm.vn/images/avt/0.png?1311)
A=5+52+...+599+5100
=(5+52)+...+(599+5100)
=5.(1+5)+...+599.(1+5)
=5.6+...+599.6
=6.(5+...+599) chia hết cho 6 (dpcm)
Ccá câu khcs bạn cứ dựa vào câu a mà làm vì cách làm tương tự chỉ hơi khác 1 chút thôi
Chúc bạn học giỏi nha!!
\(A=5+5^2+5^3+...+5^{100}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...\left(5^{99}+5^{100}\right)\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{99}.6\)
\(=6\left(5+5^3+...+5^{99}\right)⋮6\)(đpcm)
\(B=2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+...+2^{96}.31\)
\(=31\left(2+...+9^{96}\right)⋮31\)(đpcm)
\(C=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{59}.4\)
\(=4\left(3+3^3+...+3^{59}\right)⋮4\)(đpcm)
\(C=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=3.13+...+3^{58}.13\)
\(=13\left(3+...+3^{58}\right)⋮13\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:A=(1+5^1+5^2)+(5^3+5^4+5^5)+....+(5^57+5^58+5^59)
=31+31*5^3+....+31*5^57
=(1+5^3+....+5^57)*31
=>a chia hết cho 31
a)A=2+2^2+2^3+...+2^60 chia hết cho 15
=>(2+2^2+2^3+2^4)+...+(2^57+2^58+2^59+2^60)
=>2.(1+2+2^2+2^3)+...+2^57+(1+2+2^2+2^3)
=>2.15+...+2^57.15
Vì 15 chia hết choo 15
=>a chia hết cho 15
b)B=1+5+5^2+5^3+...+5^56+5^59+5^98 chia hết cho 31
=>(1+5+5^2)+...+5^56.(1+5+5^2)
=>31+....+5^56.3vi2 31 chia hết cho 31
=>B chia hết cho 31
Ta có :
=2+2^2+2^3+...+2^60 = 2(1+2+2^2+2^3) + 2^5(1+2+2^2+2^3) + ... + 2^57(1+2+2^2+2^3)
A=(2+2^5+...+2^57)*15 chia het cho 15