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ko hiểu? 

a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)

1990.1990 -1992.1988

a: Ta có: \(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=3\cdot\left(2+2^3+...+2^{99}\right)⋮3\)

b: Ta có: \(B=4+4^2+4^3+...+4^{2022}\)

\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)\)

\(=5\cdot\left(4+4^3+...+4^{2021}\right)⋮5\)

a)\(A=2+2^2+2^3+2^4+...+2^{100}=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)=3.2+3.2^3+...+3.2^{99}=3\left(2+2^3+...+2^{99}\right)⋮3\)b) \(B=4+4^2+4^3+...+4^{2022}=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)=5.4+5.4^3+...+5.4^{2021}=5\left(4+4^3+...+4^{2021}\right)⋮5\)

\(A=2+2^2+...+2^{20}\)

\(2A=2^2+2^3+...+2^{21}\)

\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)

\(A=2^{21}-2\)

___________

\(B=5+5^2+...+5^{50}\)

\(5B=5^2+5^3+...+5^{51}\)

\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)

\(4B=5^{51}-5\)

\(B=\dfrac{5^{51}-5}{4}\)

___________

\(C=1+3+3^2+...+3^{100}\)

\(3C=3+3^2+...+3^{101}\)

\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)

\(2C=3^{101}-1\)

\(C=\dfrac{3^{101}-1}{2}\)

2A= 2(2+2²+2³+...+2¹⁹+2²⁰)

2A= 2²+2³+2⁴+...+2²⁰+2²¹

2A-A=(2²+2³+2⁴+...+2²⁰+2²¹)-(2+2²+2³+...+2¹⁹+2²⁰)

A=2²¹-2

Vậy A=2²¹-2

Làm tương tự nhee

khó v

\(S=5+5^2+5^3+5^4+...+5^{2016}\\ =\left(5+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)...+\left(5^{2013}+5^{2014}+5^{2015}+5^{2016}\right)\\ =\left(5+5^2+5^3+5^4\right)+5^4\left(5+5^2+5^3+5^4\right)+...+5^{2012}\left(5+5^2+5^3+5^4\right)\\ =780+5^4\cdot780+...+5^{2012}\cdot780\\ =780\cdot\left(5^4+...+5^{2012}\right)=65\cdot12\cdot\left(5^4+...+5^{2012}\right)⋮65\)vậy S chia hết cho 65

a) \(S=1+2+2^2+..+2^{2022}\)

\(2S=2+2^2+2^3+...+2^{2023}\)

\(2S-S=2+2^2+2^3+...+2^{2023}-1-2-2^2-...-2^{2022}\)

\(S=2^{2023}-1\)

b) \(S=3+3^2+3^3+...+3^{2022}\)

\(3S=3^2+3^3+...+3^{2023}\)

\(3S-S=3^2+3^3+....+3^{2023}-3-3^2-...-3^{2022}\)

\(2S=3^{2023}-3\)

\(\Rightarrow S=\dfrac{3^{2023}-3}{2}\)

c) \(S=4+4^2+4^3+...+4^{2022}\)

\(4S=4^2+4^3+...+4^{2023}\)

\(4S-S=4^2+4^3+...+4^{2023}-4-4^2-...-4^{2022}\)

\(3S=4^{2023}-4\)

\(S=\dfrac{4^{2023}-4}{3}\)

d) \(S=5+5^2+...+5^{2022}\)

\(5S=5^2+5^3+...+5^{2023}\)

\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)

\(4S=5^{2023}-5\)

\(S=\dfrac{5^{2023}-5}{4}\)

a: \(2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+48\)

\(=51\)

b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)

\(=5\cdot\left[10+90\right]-5\)

=495

a: 23−53:52+12⋅2223−53:52+12⋅22

=8−5+48=8−5+48

=51=51

b: 5⋅[(85−35:7):8+90]−55⋅[(85−35:7):8+90]−5

=5⋅[10+90]−5=5⋅[10+90]−5

=495

Ta có A = 5 + 5² + 5³ + ... + 5²⁰²¹

5A = 5² + 5³ + 5⁴ + ... + 5²⁰²²

5A - A = ( 5² + 5³ + 5⁴ + ... + 5²⁰²² ) - ( 5 + 5² + 5³ + ... + 5²⁰²¹ )

4A = 5²⁰²² - 5

A = \(\dfrac{5^{2022}-5}{4}\)

Tìm chữ số tận cùng của kết quả mỗi phép tính sau:

a. 49¹⁵

b. 54¹⁰

c. 11²⁰+119²¹+2000²²

a) \(3.5^2+15.2^2-26\div2\)

= 3.25 + 15.4 - 13

= 75 + 60 - 13

= 135 - 13

= 122

b) \(5^3.2-100\div4+2^3.5\)

= 125.2 - 25 + 8.5

= 250 - 25 + 40

= 225 + 40

= 265

c)\(6^2\div9+50.2-3^3.33\)

= 36 : 9 + 100 - 9.33

= 4 + 100 - 297

= 104 - 297

= -193

d)\(3^2.5+2^3.10-81\div3\)

= 9.5 + 8.10 - 27

= 45 + 80 - 27

= 125 - 27

= 98

e) \(5^{13}\div5^{10}-25.2^2\)

= 5³ - 25.4

= 125 - 100

= 25

f) \(20\div2^2+5^9\div5^8\)

= 20 : 4 + 5

= 5 + 5

= 10

a.

$S=1+2+2^2+2^3+...+2^{2017}$
$2S=2+2^2+2^3+2^4+...+2^{2018}$

$\Rightarrow 2S-S=(2+2^2+2^3+2^4+...+2^{2018}) - (1+2+2^2+2^3+...+2^{2017})$

$\Rightarrow S=2^{2018}-1$

b.

$S=3+3^2+3^3+...+3^{2017}$
$3S=3^2+3^3+3^4+...+3^{2018}$

$\Rightarrow 3S-S=(3^2+3^3+3^4+...+3^{2018})-(3+3^2+3^3+...+3^{2017})$

$\Rightarrow 2S=3^{2018}-3$
$\Rightarrow S=\frac{3^{2018}-3}{2}$
 

Câu c, d bạn làm tương tự a,b. 

c. Nhân S với 4. Kết quả: $S=\frac{4^{2018}-4}{3}$

d. Nhân S với 5. Kết quả: $S=\frac{5^{2018}-5}{4}$