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![](https://rs.olm.vn/images/avt/0.png?1311)
A = 5 + 5² + 5³ + ... + 5²⁰²³
⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²⁴
⇒ 4A = 5A - A
= (5² + 5³ + 5⁴ + ... + 5²⁰²⁴) - (5 + 5² + 5³ + ... + 5²⁰²³)
= 5²⁰²⁴ - 5
⇒ A = (5²⁰²⁴ - 5)/4
A = 5 + 5² + 5³ + ... + 5²⁰²³
⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²⁴
⇒ 4A = 5A - A
= (5² + 5³ + 5⁴ + ... + 5²⁰²⁴) - (5 + 5² + 5³ + ... + 5²⁰²³)
= 5²⁰²⁴ - 5
⇒ A = (5²⁰²⁴ - 5)/4
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 5 + 5² + 5³ + ... + 5⁴⁹ + 5⁵⁰
⇒ 5A = 5² + 5³ + 5⁴ + ... + 5⁵⁰ + 5⁵¹
⇒ 4A = 5A - A
= (5² + 5³ + 5⁴ + ... + 5⁵⁰ + 5⁵¹) - (5 + 5² + 5³ + ... + 5⁴⁹ + 5⁵⁰)
= 5⁵¹ - 5
⇒ A = (5⁵¹ - 5) : 4
![](https://rs.olm.vn/images/avt/0.png?1311)
1-2+3-4+5-6+...+51-52+53
=(1-2)+(3-4)+...+(51-52)+53
=(-1)+(-1)+...+(-1)+53
=(-1)×26+53
=-26+53
=27
1-2+3-4+5-6+...+51-52+53
=(1-2)+(3-4)+...+(51-52)+53
=(-1)+(-1)+...+(-1)+53
=(-1)×26+53
=-26+53
=27
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(7\cdot\left(-2\right)^3-12\cdot\left(-5\right)+\left(-17\right)\)
\(=7\cdot\left(-8\right)+60-17\)
=-56+43
=-13
b: \(1632-37-\left(-157\right)-163-1532\)
\(=\left(1632-1532\right)-37-163+157\)
=100-200+157
=57
c: \(47\cdot\left(-918\right)+\left(-53\right)\cdot918\)
\(=918\left(-47\right)+\left(-53\right)\cdot918\)
\(=918\cdot\left(-47-53\right)\)
\(=918\left(-100\right)=-91800\)
d: \(\left(-52\right)\cdot\left(-281\right)+\left(-52\right)\cdot181\)
\(=\left(-52\right)\left(-281+181\right)\)
\(=\left(-52\right)\cdot\left(-100\right)=5200\)
a: 7⋅(−2)3−12⋅(−5)+(−17)7⋅(−2)3−12⋅(−5)+(−17)
=7⋅(−8)+60−17=7⋅(−8)+60−17
=-56+43
=-13
b: 1632−37−(−157)−163−15321632−37−(−157)−163−1532
=(1632−1532)−37−163+157=(1632−1532)−37−163+157
=100-200+157
=57
c: 47⋅(−918)+(−53)⋅91847⋅(−918)+(−53)⋅918
=918(−47)+(−53)⋅918=918(−47)+(−53)⋅918
=918⋅(−47−53)=918⋅(−47−53)
=918(−100)=−91800=918(−100)=−91800
d: (−52)⋅(−281)+(−52)⋅181(−52)⋅(−281)+(−52)⋅181
=(−52)(−281+181)=(−52)(−281+181)
=(−52)⋅(−100)=5200=(−52)⋅(−100)=5200
![](https://rs.olm.vn/images/avt/0.png?1311)
5A=5+5^2+...+5^2023
=>4A=5^2023-1
=>\(A=\dfrac{5^{2023}-1}{4}\)
\(2B-A=\dfrac{5^{2023}}{4}-\dfrac{5^{2023}-1}{4}=\dfrac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S=5+5^2+...+5^{2006}\)
\(5S=5^2+5^3+...+5^{2007}\)
\(5S-S=5^2+5^3+5^4+...+5^{2007}-5-5^2-5^3-...-5^{2006}\)
\(4S=5^{2007}-5\)
\(S=\dfrac{5^{2007}-5}{4}\)
b) \(S=5+5^2+5^3+...+5^{2006}\)
\(S=\left(5+5^4\right)+\left(5^2+5^5\right)+...+\left(5^{2003}+5^{2006}\right)\)
\(S=5\cdot\left(1+5^3\right)+5^2\cdot\left(1+5^3\right)+...+5^{2003}\cdot\left(1+5^3\right)\)
\(S=\left(1+5^3\right)\cdot\left(5+5^2+...+5^{2003}\right)\)
\(S=126\cdot\left(5+5^2+...+5^{2003}\right)\) ⋮ 126
![](https://rs.olm.vn/images/avt/0.png?1311)
=(1-2)+(3-4)+...+(53-54)
=-1+(-1)+(-1)+...+(-1)
(dãy có (54-1):1+1=54 số hạng
=> dãy có 27 cặp)
=27.(-1)=-27
#Học-tốt
1-2+3-4+5-6+........+51-52+53-54 (54 Số Hạng)
= (1-2)+(3-4)+(5-6)+.......+(51-52)+(53-54) ( 54 : 2 = 27 Nhóm)
= (-1)+(-1)+(-1)+........+(-1)+(-1) (27 Số Hạng)
= (-1) x 27
= (-27)
Bài Này Dễ Mà
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 5A=5+5^2+...+5^2023
=>4A=5^2023-1
=>A=(5^2023-1)/4
b: 6B=6^2+6^3+...+6^41
=>5B=6^41-6
=>B=(6^41-6)/5
c: 16C=4^4+4^6+...+4^16
=>15C=4^16-4^2
=>C=(4^16-4^2)/15
d: 9D=3^3+3^5+...+3^27
=>8D=3^27-3
=>D=(3^27-3)/8
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(S=1-5+5^2-5^3+...+5^{98}-5^{99}\)
=>\(5S=5-5^2+5^3-5^4+...+5^{99}-5^{100}\)
=>\(6S=5-5^2+5^3-5^4+...+5^{99}-5^{100}+1-5+5^2-5^3+...+5^{98}-5^{99}\)
=>\(6S=-5^{100}+1\)
=>\(S=\dfrac{-5^{100}+1}{6}\)
b: S=1-5+52-53+...+598-599 là số nguyên
=>\(\dfrac{-5^{100}+1}{6}\in Z\)
=>\(-5^{100}+1⋮6\)
=>\(5^{100}-1⋮6\)
=>\(5^{100}\) chia 6 dư 1
Đặt \(T=5+5^2+5^3+...+5^{2000}\)
\(5T=5^2+5^3+5^4+...+5^{2001}\)
Mà \(5T-T=4T=5^{2001}-5\Rightarrow T=\frac{5^{2001}-5}{4}\)