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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: \(2022\cdot2023^{2024}+2023^{2024}=2023^{2024}\left(2022+1\right)\)
\(=2023^{2025}\)
b) 32, 33, 34 , 35.
\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)
\(b,3^2=9,3^3=27,3^4=81,3^5=243\)
\(c,4^2=16,4^3=64,4^4=256\)
\(d,5^2=25,5^3=125,5^4=625\)
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: $S=\frac{5^{2018}-5}{4}$
1.
a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)
\(2A=2+2^2+2^3+....+2^{2008}\)
b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)
\(=2^{2008}-1\) (bạn xem lại đề)
2.
\(A=1+3+3^1+3^2+...+3^7\)
a. \(2A=2+2.3+2.3^2+...+2.3^7\)
b.\(3A=3+3^2+3^3+...+3^8\)
\(2A=3^8-1\)
\(=>A=\dfrac{2^8-1}{2}\)
3
.\(B=1+3+3^2+..+3^{2006}\)
a. \(3B=3+3^2+3^3+...+3^{2007}\)
b. \(3B-B=2^{2007}-1\)
\(B=\dfrac{2^{2007}-1}{2}\)
4.
Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)
a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)
b.\(4C-C=4^7-1\)
\(C=\dfrac{4^7-1}{3}\)
5.
\(S=1+2+2^2+2^3+...+2^{2017}\)
\(2S=2+2^2+2^3+2^4+...+2^{2018}\)
\(S=2^{2018}-1\)
4:
a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6
=>4*C=4+4^2+...+4^7
b: 4*C=4+4^2+...+4^7
C=1+4+...+4^6
=>3C=4^7-1
=>\(C=\dfrac{4^7-1}{3}\)
5:
2S=2+2^2+2^3+...+2^2018
=>2S-S=2^2018-1
=>S=2^2018-1
\(A=2+2^2+...+2^{20}\)
\(2A=2^2+2^3+...+2^{21}\)
\(2A-A=2^2+2^3+...+2^{21}-2-2^2-...-2^{20}\)
\(A=2^{21}-2\)
___________
\(B=5+5^2+...+5^{50}\)
\(5B=5^2+5^3+...+5^{51}\)
\(5B-B=5^2+5^3+...+5^{51}-5-5^2-...-5^{50}\)
\(4B=5^{51}-5\)
\(B=\dfrac{5^{51}-5}{4}\)
___________
\(C=1+3+3^2+...+3^{100}\)
\(3C=3+3^2+...+3^{101}\)
\(3C-C=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}\)
\(2C=3^{101}-1\)
\(C=\dfrac{3^{101}-1}{2}\)
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
a: \(2A=2^1+2^2+...+2^{2022}\)
\(\Leftrightarrow A=2^{2022}-1\)
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) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
A = 1 + 2 + 22 + ... + 2100
=> 2A = 2 + 22 + 23 + ... + 2100 + 2101
=> 2A - A = ( 2 + 22 + 23 + ... + 2100 + 2101 ) - ( 1 + 2 + 22 + ... + 2100 )
=> A = 2101 - 1
A = 1 + 2 +22+.....+2100
=> 2A =2 + 22 + 23+...+2100+2101
=> 2A - A = ( 2 + 22+23+.....+2100+2101) - ( 1 + 2 + 22+...+2100)
=> A = 2101 - 1