C = 4 + 42 + 43 + ...... + 4n
D = 1 + 5 + 52 + ...... + 52000
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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: $S=\frac{5^{2018}-5}{4}$
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)\(...A=\dfrac{2^{50+1}-1}{2-1}=2^{51}-1\)
b) \(...\Rightarrow B=\dfrac{3^{80+1}-1}{3-1}=\dfrac{3^{81}-1}{2}\)
c) \(...\Rightarrow C+1=1+4+4^2+4^3+...+4^{49}\)
\(\Rightarrow C+1=\dfrac{4^{49+1}-1}{4-1}=\dfrac{4^{50}-1}{3}\)
\(\Rightarrow C=\dfrac{4^{50}-1}{3}-1=\dfrac{4^{50}-4}{3}=\dfrac{4\left(4^{49}-1\right)}{3}\)
Tương tự câu d,e,f bạn tự làm nhé
1 < 2
3 > 1
3 < 4
3 = 3
5 > 2
5 > 4
2 < 3
1 < 5
4 > 1
4 = 4
4 > 3
5 = 5
2 < 3
3 < 5
1 < 4
3 > 1
a: \(12+2^2+3^2+4^2+5^2\)
\(=12+4+9+16+25\)
\(=16+50=66\)
\(\left(1+2+3+4+5\right)^2=15^2=225\)
=>\(12+2^2+3^2+4^2+5^2< \left(1+2+3+4+5\right)^2\)
b: \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2< \left(1+2+3+4\right)^3\)
c: \(5^{202}=5^2\cdot5^{200}=25\cdot5^{200}>16\cdot5^{200}\)
d: \(18\cdot4^{500}=18\cdot2^{1000}\)
\(2^{1004}=2^4\cdot2^{1000}=16\cdot2^{1000}\)
=>\(18\cdot4^{500}>2^{1004}\)
e: \(2022\cdot2023^{2024}+2023^{2024}=2023^{2024}\left(2022+1\right)\)
\(=2023^{2025}\)
\(T=5+5^2+5^3+...+5^{2000}\)
=>\(5T=5^2+5^3+5^4+...+5^{2001}\)
=>\(5T-T=5^2+5^3+...+5^{2001}-5-5^2-...-5^{2000}\)
=>\(4T=5^{2001}-5\)
=>\(4T+5=5^{2001}\)
Sửa đề:\(4T+5=5^m\)
=>\(5^m=5^{2001}\)
=>m=2001
T=5+52+53+...+52000
=>5T=52+53+54+...+52001
=>5T−T=52+53+...+52001−5−52−...−52000
=>4T=52001−5
=>4T+5=52001
Ta có:4T+5=5m
=>52001=5m
=>m=2001
Vậy m=2001
a, X = 273
b, X = 156
c, X = 4
d, X = 505
e, X = \(\frac{37}{245}\)
\(27.332+93.43+57.61+69.57\\ =27.332+93.43+57.\left(61+69\right)\\ =27.332+93.43+57.130\\ =8964+3999+7410=20373\\ 34.75+75.66-65.100\\ =\left(34+66\right).75-65.100\\ =100.75-65.100\\ =100.\left(75-65\right)\\ =100.10=1000\\ \left(456.11+912\right).37:13:74\\ =5928:13:\left(74:37\right)\\ =456:2=228\\ 6^2:4.3+2.5^2\\ =36:4.3+2.25\\ =9.3+50=27+50=77\\ 5.4^2-18:3^2\\ =5.16-18:9\\ =80-2=78\\\left[\left(315+372\right).3+\left(372+315\right).7\right]:\left(26.13+74.14\right)\\ =687.\left(3+7\right):\left(338+1036\right)\\ 687.10:1374\\ =6870:1374=5\\ 12:\left\{390:\left[500-\left(125+35.7\right)\right]\right\}\\ =12:\left[390:\left(500-370\right)\right]\\ =12:\left(390:130\right)=12:3=4\\ 192000-\left(1500.2+1800.3+1800.2:3\right)\\ =192000-\left(3000+5400+1200\right)\\ =192000-9600=182400\)
C = 4 + 42 + 43 + ... + 4n
4C = 42 + 43 + 44 + ... + 4n+1
4C - C = (42 + 43 + 44 + ... + 4n+1) - (4 + 42 + 43 + ... + 4n)
3C = 4n+1 - 4
C = 4n+1 - 4/3
Câu còn lại lm tương tự