CM:
M= 1/5 + (1/5)^2 +(1/5)^3+....+(1/5)^50 < 1/4
Cho biểu thức : M = 1/5+(1/5)^2+(1/5)^3+.....+(1/5)^49+(1/5)^50
CMR M<1/4
làm đi
=>5M=1+1/5+1/5^2+...+1/5^48+1/5^49
=>5M-M=(1+1/5+1/5^2+..+1/5^48+1/5^49)-(1/5+1/5^2+1/5^3+...+1/5^49+1/5^50)
=>4M=1-1/5^50
=>M=(1-1/5^50)/4
mà 1-1/5^50<1
=>M<1/4(đpcm)
C/m
1/5+(1/5)^2+(1/5)^3+...+(1/5)^50<1/4
Đặt \(A=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)
\(5.A=1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\)
\(5.A-A=\left(1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\right)-\left(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\right)\)
\(4.A=1-\frac{1}{5^{50}}< 1\)
\(\Rightarrow A< \frac{1}{4}\)
giup voi a minh dang can gap ai nhanh minh cho dung nhe
A=2 mũ 3 + 2 mũ 4 + 2 mũ 5 + 2 mũ 6 + 2 mũ 7 +.....+ 2 mũ 90
B=1+5+5 mũ 2 + 5 mũ 3 +5 mũ 4 +......+5 mũ 50
C=1/5 +1/5 mũ 2 + 1/5 mũ 3 + 1/5 mũ 4 +1/5 mũ 6 +......+1/5 mũ 102
D=1/5 +1/5 mũ 3 + 1/5 mũ 4 +1/5 mũ 5 + 1/5 mũ 6 +1/5 mũ 105
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
bai 1
1) (-3)^10 . 15^5 / 25^3 . (-9)^7
2) (-5)^60 . 30^5 / 15^5 . 5 ^61
3) 3/5+ 3/5^3 + 3/5^5 + ... + 3/5^ 199
4) 3^6 . 45^4 - 15^13 . 5^-9 / 27^4 . 25^3 + 45^6
bai 2 ; so sanh
1) (-5)^30 ; (-3) ^50
2) ( -1/16)^10 ; (1/2)^50
3) 2^2010+1 / 2^2007 + 1 ; 2^2012+1 / 2^2009+1
Cho A = 1/3^2 + 1/4^2 + 1/5^2 + ... + 1/50^2
CM rằng A > 1/4
Ta có : \(A>\frac{1}{3^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}\)
\(\rightarrow A>\frac{1}{9}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{50}-\frac{1}{51}\)
\(\rightarrow A>\frac{1}{4}+\left(\frac{1}{9}-\frac{1}{51}\right)\)
Xét : \(\frac{1}{9}-\frac{1}{51}>0\rightarrow A>\frac{1}{4}\left(đpcm\right)\)
Tính
A=13/21.2/11+13/21.9/11+8/21.
B=(1-1/5).(1-2/5).(1-3/5)...(1-9/5)
C= (1-1/2).(1-1/3).(1-1/4)...(1-1/50)
D= 2²/1.3 . 3²/2.4 . 4²/3.5 . 5²/4.6 . 6²/5.7
A = 13/21.2/11 + 13/21.9/11 + 8/21
= (13/21) + (13/21) + (8/21)
= (13 + 13 + 8)/21
= 34/21
B = (1 - 1/5)(1 - 2/5)(1 - 3/5)...(1 - 9/5)
= (4/5)(3/5)(2/5)(1/5)(0/5)(-1/5)(-2/5)(-3/5)(-4/5)
= 0
C = (1 - 1/2)(1 - 1/3)(1 - 1/4)...(1 - 1/50)
= (1/2)(2/3)(3/4)(4/5)...(49/50)
= 1/50
D = (2^2/1.3) * (3^2/2.4) * (4^2/3.5) * (5^2/4.6) * (6^2/5.7)
= (4/3) * (9/8) * (16/15) * (25/23) * (36/35)
= 0.979
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5)...+1/(1+2+3+4+5...+99)+1/50 là:
Đặt \(S=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+.....+99}+\frac{1}{50}\)
Đặt E = \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+....+99}\)
\(E=\frac{1}{2.3:2}+\frac{1}{3.4:2}+....+\frac{1}{99.100:2}\)
\(\frac{1}{2}E=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
E = 49/100 : 1/2 = 49/50
Vậy \(S=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
1/1*2+1/3*4+1/5*6+...+1/49*50=1/26+1/27+1/28+...+1/50
1/1*2+1/3*4+1/5*6+...+1/99*100