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Bài 1:
$B=1+3+3^2+3^3+...+3^{100}$
$=1+(3+3^2)+(3^3+3^4)+...+(3^{99}+3^{100})$
$=1+3(1+3)+3^3(1+3)+...+3^{99}(1+3)$
$=1+(1+3)(3+3^3+...+3^{99})=1+4(3+3^3+....+3^{99})$
$\Rightarrow B$ chia 4 dư 1.
Bài 2:
$C=5-5^2+5^3-5^4+...+5^{2023}-5^{2024}$
$5C=5^2-5^3+5^4-5^5+...+5^{2024}-5^{2025}$
$\Rightarrow C+5C=5-5^{2025}$
$6C=5-5^{2025}$
$C=\frac{5-5^{2025}}{6}$
a: \(24\cdot35+65\cdot24-1400\)
\(=24\left(35+65\right)-1400\)
\(=24\cdot100-1400=2400-1400=1000\)
b: \(32:2^4-4\cdot5^2\)
\(=\dfrac{32}{16}-4\cdot25\)
=2-100
=-98
c: \(17+75:\left[30-5\cdot\left(7^2-48\right)\right]\)
\(=17+75:\left[30-5\cdot\left(49-48\right)\right]\)
\(=17+75:\left[30-5\right]\)
\(=17+\dfrac{75}{25}=17+3=20\)
d: \(\left(-138\right)-\left[258+4\cdot\left(-36\right)\right]\cdot2024^0\)
\(=-138-\left[258-144\right]\)
\(=-138-114\)
=-252
a: \(61\cdot45+61\cdot23-68\cdot51\)
\(=61\left(45+23\right)-68\cdot51\)
\(=68\cdot61-68\cdot51\)
\(=68\left(61-51\right)=68\cdot10=680\)
b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)
\(=75-75+4\cdot8\)
\(=4\cdot8=32\)
c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)
\(=\dfrac{36}{20-30+4^2}\)
\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)
d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)
\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)
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
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) \(S=5+5^2+...+5^{2022}\)
\(5S=5^2+5^3+...+5^{2023}\)
\(5S-S=5^2+5^3+...+5^{2023}-5-5^2-...-5^{2022}\)
\(4S=5^{2023}-5\)
\(S=\dfrac{5^{2023}-5}{4}\)
22 x 43 - 625 : {[504 - (52 x 8 + 70) : 32 + 6] : 20 + 20230}
= 4 x 64 - 625 : {[504 - (25 x 8 + 70) : 9 + 6] : 20 + 1}
= 256 - 625 : {[504 - (200 + 70) : 9 + 6] : 20 + 1}
= 256 - 625 : {[504 - 270 : 9 + 6] : 20 + 1}
= 256 - 625 : {[504 - 30 + 6] : 20 + 1}
= 256 - 625 : {[474 + 6] : 20 + 1]
= 256 - 625 : {480 : 20 + 1]
= 256 - 625 : {24 + 1}
= 256 - 625 : 25
= 256 - 41
= 215
22 x 43 - 625 : {[504 - (52 x 8 + 70) : 32 + 6] : 20 + 20230}
= 4 x 64 - 625 : {[504 - (25 x 8 + 70) : 9 + 6] : 20 + 1}
= 256 - 625 : {[504 - (200 + 70) : 9 + 6] : 20 + 1}
= 256 - 625 : {[504 - 270 : 9 + 6] : 20 + 1}
= 256 - 625 : {[504 - 30 + 6] : 20 + 1}
= 256 - 625 : {[474 + 6] : 20 + 1]
= 256 - 625 : {480 : 20 + 1]
= 256 - 625 : {24 + 1}
= 256 - 625 : 25
= 256 - 41
= 215
Ta có :
A = 1 + 5 + \(5^2\)+\(5^3\)+...+ \(5^{2023}\)
5A = 5 + \(5^2\)+\(5^3\)+\(5^4\)+..+ \(5^{2024}\)
=> 5A - A = ( 5 + \(5^2\)+\(5^3\)+\(5^4\)+..+ \(5^{2024}\) ) - ( 1 + 5 + \(5^2\)+\(5^3\)+...+ \(5^{2023}\) )
=> 4A = \(5^{2024}\)- 1
Nhận thấy :
\(5^{2024}\) - 1 > \(5^{2024}\)
=> 4A < \(5^{2024}\)
Vậy 4A < \(5^{2024}\)
a) 23.25 + 75.23 - 1300
= 23.(25 + 75) - 1300
= 23.100 - 1300
= 2300 - 1300
= 1000
b) 36 : 3² - 5.2²
= 36 : 9 - 5.4
= 4 - 20
= -16
c) 183 + 80 : [20 - 4.(5² - 24)]
= 183 + 80 : [20 - 4.(25 - 24)]
= 183 + 80 : (20 - 4.1)
= 183 + 80 : 16
= 183 + 5
= 188
d) (-125) - [148 + 5.(-25)]
= -125 - (148 - 125)
= -125 - 148 + 125
= (-125 + 125) - 148
= 0 - 148
= -148
a) 23.25 + 75.23 - 1300
= 23. (25+75) - 1300
= 23.100 - 1300
= 2300 - 1300 = 1000
b) 36 : 32 - 5 . 22
= 36: 9 - 5.4
= 4 - 20 = -16
c) 183 + 80 : [ 20 - 4 ( 52 - 24 ) ]
= 183 + 80: [20 - 4.(25-24)]
= 183 + 80 : [20 - 4.1)]
= 183 + 80: [20 - 4]
= 183 + 80:16
= 183 + 5 = 188
d) (-125) - [148 + 5 . (-25) ]
= -125 - 148 + 125
= (125 - 125) - 148
= 0 - 148 = -148
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: $S=\frac{5^{2018}-5}{4}$
\(=60-\dfrac{\left[36+64\right]}{25}\)
\(=60-4=56\)
\(60-[(36+4^3):5^2]\\=60-[(36+64):25]\\=60-(100:25)\\=60-4\\=56\)