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![](https://rs.olm.vn/images/avt/0.png?1311)
S = 5 + 52 + 53 + 54 + .......... + 599
a) S = ( 5 + 52 + 53 ) + ( 54 + 55 + 56 ) + .... + ( 597 + 598 + 599 )
= 5. ( 1 + 5 + 52 ) + 54 . ( 1 + 5 + 52 ) + .... + 597 . ( 1 + 5 + 52 )
= ( 1 + 5 + 52 ). ( 5 + 54 + .. + 597 )
= 31 . ( 5 + 54 + .... + 597 ) chia hết cho 31 ( đpcm )
c ) 5S = 52 + 53 + .. + 5100
=> 5S - S = 4S = 5100 + 599 + ........ + 53 + 52 - 5 - 52 - 53 - ..... - 599
= 5100 - 5
25x - 5 = 4S
=> 25x - 5 = 5100 - 5
=> 25x = 5100
=> 25x = ( 52 )50
=> 25x = 2550
=> x = 50
Vậy x = 50
Câu b quên cách làm rồi
a) S=5+52+53+54+...+599
=(5+52+53)+(54+55+56)+...+(597+598+599)
=5(1+5+52)+54(1+5+52)+...+597(1+5+52)
=5.31+54.31+...+597.31
=31(5+54+...+597)⋮31(đpcm)
b) S=5+52+53+54+...+599
=5+(52+53)+(54+55)+...+(598+599)
=5+5(5+52)+53(5+52)+...+597(5+52)
=5+5.30+53.30+...+597.30
=5+30.(5+53+...+597)
Mà 5⋮̸30 nên S⋮̸30(đpcm)
c) Ta có: 5S=52+53+54+55+...+5100
5S−S=(52+53+54+55+...+5100)−(5+52+53+54+...+599)
4S=5100−5
⇒25x−5=5100−5
⇒25x=5100
⇒25x=2550
⇒x=50
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 5⁰ + 5¹ + 5² + ... + 5²⁰²³
= (5⁰ + 5¹) + (5² + 5³) + ... + (5²⁰²² + 5²⁰²³)
= 6 + 5².(1 + 5) + ... + 5²⁰²².(1 + 5)
= 6 + 5².6 + ... + 5²⁰²².6
= 6.(1 + 5² + ... + 5²⁰²²) ⋮ 6
Vậy S ⋮ 6
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Số số hạng của S:
2023 - 0 + 1 = 2024 (số)
2024 : 3 dư 2 nên khi nhóm các số hạng của S theo nhóm 3 thì dư 2 số hạng
Ta có:
S = 5⁰ + 5¹ + 5² + 5³ + ... + 5²⁰²³
= 5⁰ + 5¹ + (5² + 5³ + 5⁴) + (5⁵ + 5⁶ + 5⁷) + ... + (5²⁰²¹ + 5²⁰²² + 5²⁰²³)
= 6 + 5².(1 + 5 + 5²) + 5⁵.(1 + 5 + 5²) + ... + 5²⁰²¹.(1 + 5 + 5²)
= 6 + 5².31 + 5⁵.31 + ... + 5²⁰²¹.31
= 6 + 31.(5² + 5⁵ + ... + 5²⁰²¹)
Do 31.(5² + 5⁵ + ... + 5²⁰²¹) ⋮ 31
6 + 31.(5² + 5⁵ + ... + 5²⁰²¹) chia 31 dư 6
Vậy S chia 31 dư 6
------------
Sửa đề:
Tìm số tự nhiên n để 4S - 25² = -1
S = 5⁰ + 5¹ + 5² + 5³ + ... + 5²⁰²³
5S = 5 + 5² + 5³ + 5⁴ + ... + 5²⁰²⁴
⇒ 4S = 5S - S
= (5 + 5² + 5² + 5³ + ... + 5²⁰²⁴) - (1 + 5¹ + 5² + 5³ + ... + 5²⁰²³)
= 5²⁰²⁴ - 1
⇒ 4S - 25²ⁿ = -1
⇒ 5²⁰²⁴ - 1 - (5²)²ⁿ = -1
⇒ 5²⁰²⁴ - 5⁴ⁿ = -1 + 1
⇒ 5⁴ⁿ = 5²⁰²⁴
⇒ 4n = 2024
⇒ n = 2024 : 4
⇒ n = 506
\(S=\left(5^0+5^1\right)+\left(5^2+5^3\right)+...+\left(5^{2022}+5^{2023}\right)\\ =6+5^2\left(1+5\right)+...+5^{2022}\left(1+5\right)\\ =6+5^2.6+...+5^{2022}.6\\ =6\left(1+5^2+...+5^{2022}\right)⋮6\)
\(S=\left(5^0+5^1+5^2\right)+...+\left(5^{2021}+5^{2022}+5^{2023}\right)\\ =31+...+5^{2021}\left(1+5+5^2\right)\\ =31\left(1+...+5^{2021}\right)⋮31\)
=> Dư : 0
\(5S=5^1+5^2+5^3+5^4+...+5^{2024}\\ =>5S-S=4S=5^{2024}-1\)
Mà : \(4S-25^{2n}=1\\ =>5^{2024}-1-25^{2n}=1\\ =>5^{2024}-25^{2n}=2\)
Bạn xem lại đề nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S=5+5^2+5^3+5^4+...+5^{99}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{97}+5^{98}+5^{99}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{97}\left(1+5+5^2\right)\)
\(=5.31+5^4.31+...+5^{97}.31\)
\(=31\left(5+5^4+...+5^{97}\right)⋮31\left(đpcm\right)\)
b) \(S=5+5^2+5^3+5^4+...+5^{99}\)
\(=5+\left(5^2+5^3\right)+\left(5^4+5^5\right)+...+\left(5^{98}+5^{99}\right)\)
\(=5+5\left(5+5^2\right)+5^3\left(5+5^2\right)+...+5^{97}\left(5+5^2\right)\)
\(=5+5.30+5^3.30+...+5^{97}.30\)
\(=5+30.\left(5+5^3+...+5^{97}\right)\)
Mà \(5⋮̸30\) nên \(S⋮̸30\left(đpcm\right)\)
c) Ta có: \(5S=5^2+5^3+5^4+5^5+...+5^{100}\)
\(5S-S=\left(5^2+5^3+5^4+5^5+...+5^{100}\right)-\left(5+5^2+5^3+5^4+...+5^{99}\right)\)
\(4S=5^{100}-5\)
\(\Rightarrow25^x-5=5^{100}-5\)
\(\Rightarrow25^x=5^{100}\)
\(\Rightarrow25^x=25^{50}\)
\(\Rightarrow x=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3A=3^2+3^3+...+3^{101}\\ \Rightarrow3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\\ \Rightarrow2A=3^{101}-3\\ \Rightarrow A=\dfrac{3^{101}-3}{2}\)
\(b,A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\\ A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\\ A=\left(1+3\right)\left(3+3^3+...+3^{99}\right)\\ A=4\left(3+3^3+...+3^{99}\right)⋮4\)
\(A=3+\left(3^2+3^3+...+3^{100}\right)\\ A=3+3^2\left(1+3+...+3^{100}\right)\\ A=3+9\left(1+3+...+3^{100}\right).chia.9.dư.3\\ \Rightarrow A⋮̸9\)
a) rút gọn a
a = 3 + 3^3 + 3^2 + .. + 3^100
3a = 3^2 + 3^3 + .. + 3^101
3a - a = (3^2 + 3^3 + .. + 3^101) - (3 + 3^2 + .. + 3^100)
2a = 3^301 - 3
a = 3^101 - 3/2
b) chứng minh a chia hết cho 4 và k chia hết cho 9
a = 3 + 3^2 + .. + 3^100
a = (3 + 3^2) + .. + (3^99 + 3^100)
a = 3 (1 + 3) + .. + 3^99 (1 + 3)
a = 3.4 + .. + 3^99.4
a = (3 + .. + 3^99).4 ⋮ 4
vì 9 ⋮̸4
=> a ⋮̸9
![](https://rs.olm.vn/images/avt/0.png?1311)
a là x và y thuộc nhóm rỗng
b thì =-1+-1+-1+...+-1+2017=-1008+2017=1009
c là vì 4S+1 là 5^2016 chia hết cho 5^2016
vì 6(5+5^2+...+5^2014) chia hết cho 6 và bằng S
![](https://rs.olm.vn/images/avt/0.png?1311)
S = (5 + 52) + (53 + 54) +....+(59 + 510)
S = 1.30 + 52.30+....+58.30
S = 30.(1+52+....+58)
S chia hết cho 30
=> ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
S=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9+5^10
=>S=(5+5^2)+(5^3+5^4)+(5^5+5^6)+(5^7+5^8)+(5^9+5^10)
=>S=30+5^2(5+5^2)+5^4(5+5^2)+5^6(5+5^2)+5^8(5+5^2)
=>S=30+5^2.30+5^4.30+5^6.30+5^8.30
=>S=30(1+5^2+5^4+5^6+5^8)=> S chia hết cho 30
\(5+5^2+5^3+5^4+5^5+...+5^9+5^{10}\)
\(=5+5^2+5^2\left(5+5^2\right)+5^4\left(5+5^2\right)+...+5^8\left(5+5^2\right)\)
\(=\left(5+5^2\right)\left(1+5^2+5^4+5^6+5^8\right)\)
\(=30.\left(1+5^2+5^4+5^6+5^8\right)\)
vậy S chia hết cho 30
ko hiểu họi lại mik
tick mik nka
a) \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5S=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5S-S=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(\Rightarrow4S=5^{101}-5\)
\(\Rightarrow S=\frac{5^{101}-5}{4}\)
b) \(4S+5=5^x\)
\(\Rightarrow5^{101}-5+5=5^x\)
\(\Rightarrow5^{101}=5^x\)
\(\Rightarrow x=101\)
Vậy x = 101
c) \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow S=\left(5+25\right)+5^2.\left(5+5^2\right)+...+5^{98}.\left(5+5^2\right)\)
\(\Rightarrow S=30+5^2.30+...+5^{98}.30\)
\(\Rightarrow S=\left(1+5^2+...+5^{98}\right).30⋮30\)
\(\Rightarrow S⋮30\left(đpcm\right)\)
khoai vừa S chia hết 31 thím ạ