A=1/1*2+1/2*3+1/3*4+......+1/99*100 so sanh voi 1
so sanh A va B voi 0
A = 1. ( - 2 ) . 3 . ( - 4 ) ..... . 99 . ( -100 )
B = 1 . ( - 2 ) . 3 . ( - 4 ) . ... . (-98) . 99
A có 50 thừa số âm
=> A > 0
b) CÓ 49 thừa số âm
=> B < 0
A có 50 thừa số âm
=> A > 0
B có 49 thừa số âm
=> B < 0
tick nha
so sanh A va B voi 0
A = 1 . ( - 2 ) . 3 .( - 4 ) . .... . 99 . ( - 100 )
B = 1 . ( - 2 ) . 3. ( - 4 ) . .... . ( - 98 ) . 99
cho A=(1/2^2-1).(1/3^2-1).(1/4^2-1).....(1/100^2-1). So sanh A voi 1/2
A có : 100 - 2 + 1 = 99 thừa số.
Tất cả thừa số của A đều âm.
=> A < 0 < \(\frac{1}{2}\)
cho A = (1/2^2-1)(1/3^2-1)(1/4^2-1)...(1/100^2-1). so sanh voi -1/2
So sanh A voi 1:
A=1/2*2 + 1/3*3 + 1/4*4 + .....+1/2011*2011
So sanh B voi 3/4:
B=1/2*2 + 1/3*3 +1/4*4 + ......+1/2011*2011
1/51+1/52+1/53+...1/99+1/100 hay so sanh voi 1/2
A=1/4+1/42+1/43+.....+1/499
a) Rut gon A
b) So sanh A voi 1/3
4A=1+1/4+1/42+......+1/498
4A - A = ( 1+1/4+1/42+..........+1/498) - ( 1/4+1/42+1/43+.......+1/499)
3A= 1-1/499
A= 1/3 - 1/499 : 3
Mà 1/499 : 3 > 0 => 1/3 - 1/499 : 3 < 1/3
Hay A < 1/3
a/ Rút gọn:
\(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{99}}.\)
=> \(4A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}\)
=> \(4A=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}+\frac{1}{4^{99}}\right)-\frac{1}{4^{99}}\)
<=> \(4A=1+A-\frac{1}{4^{99}}\)
=> \(3A=1-\frac{1}{4^{99}}\)
=> \(A=\frac{1}{3}-\frac{1}{3.4^{99}}\)
b/ Ta có: \(A=\frac{1}{3}-\frac{1}{3.4^{99}}< \frac{1}{3}\)
Ta có:
A=1/4+1/42+…+1/499
4A=4.(1/4+1/42+…+1/499)
4A=1+1/4+…+1/498
4A-A=(1+1/4+…+1/498)- (1/4+1/42+…+1/499)
3A=1-1/499
A=1/3-1/(499.3)
b)Vì A=1/3-1/(499.3) nên A<1/3 (do 1/499.3>0)
A=(1/2^2-1) * (1/3^2-1) *...*(1/100^2-1) so sanh A voi 2
so sanh A voi 1/2 nhe, khong phai A voi 2 dau
A=(1/2^2-1) * (1/3^2-1) *...*(1/100^2-1) so sanh A voi 1/2