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Ta thấy:
\(\frac{1}{2^2}<\frac{1}{1.2}\)
\(\frac{1}{3^2}<\frac{1}{2.3}\)
................
\(\frac{1}{19^2}<\frac{1}{18.19}\)
Cộng vế với vế ta có:
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{19^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{18.19}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}\)\(=1-\frac{1}{19}=\frac{18}{19}>\frac{18}{40}=\frac{9}{20}\)
Kết luận: ....>.....
Mình giải thử xem đúng ko nha:
M=1/22+1/32+1/42+....+1/202
=1/2.2+1/3.3+1/4.4+.....+1/20.20
mà M<1/1.2+1/2.3+1/3.4+....+1/19.20
M<1/1-1/2+1/2-1/3+...+1/19-1/20
M<1-1/20
M<19/20<1
=> M<1
1 A= 2^2+2^2+2^3+...+2^20
A= 2*2^2+2^3+...+2^20
A=2^3+2^3+...+2^20
tương tự vậy A=2^21 ( cố hiểu làm hơi tắt)
a/ 40^20=40^2.10=1600^10
3^30=3^3.10=27^10
vì 1600^10>27^10 nên 40^20>3^30
a) 40^20=(4^2)^10=16^10
30^30=(3^3)^10=27610
Vì 16<27=>16^10<27^10 hay 4^20<3^30
b) mk chịu
c) Đặt A= 1/3+1/3^2+1/3^3+...+1/3^99
=>3A=3( 1/3+1/3^2+1/3^3+...+1/3^99)
=>3A=1+1/3+1/3^2+...+1/3^98
=>3A-A=(1+1/3+1/3^2+...+1/3^98)-(1/3+1/3^2+1/3^3+...+1/3^99)
=>2A=1-1/3^99
=>A=(1-1/3^99)/2
=>A=1/2 - (1/3^99)/2 < 1/2=>a<1/2
A=20+21+22+23+24+...+220
2A=21+22+23+24+25+...+221
A=2A - A = (21+22+23+24+25+...+221) -(20+21+22+23+24+...+220)
A=221-20
A=221-1
=>A < 221
1,2 dễ ko làm
3,
S = 1 + 2 + 22 + 23 + ... + 29
2S = 2 + 22 + 23 + 24 + ... + 210
2S - S = ( 2 + 22 + 23 + 24 + ... + 210 ) - ( 1 + 2 + 22 + 23 + ... + 29 )
S = 210 - 1
Mà 5 . 28 = ( 1 + 22 ) . 28 = 28 + 210 > 210 > 210 - 1
Vậy S < 5 . 28
P = 1 + 3 + 32 + 33 + ... + 320
3P = 3 + 32 + 33 + 34 + ... + 321
3P - P = ( 3 + 32 + 33 + 34 + ... + 321 ) - ( 1 + 3 + 32 + 33 + ... + 320 )
2P = 321 - 1
P = ( 321 - 1 ) : 2 < 321
Vậy P < 321
\(3,1+5^2+5^4+...+5^{26}\)
\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)
\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)
\(=26+5^4.26+...+5^{24}.26\)
\(=26\left(5^4+...+5^{24}\right)\)
Vì \(26⋮26\)
\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)
\(4,1+2^2+2^4+...+2^{100}\)
\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)
\(=21+2^6.21...+2^{98}.21\)
\(=21\left(2^6+...+2^{98}\right)\)
Có : \(21\left(2^6+...+2^{98}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)