3)
C= (1/2).(3/4).(5/6).....(199/200).
C= (1.3.5….199)/(2.4.6…200)
C²= 1².3².5²….199²/(2².4².6²…200²)
Ta có: k²>k²-1=(k-1)(k+1) nên 2²>1.3; 4²>3.5 … 200²>199.201.
=>
C² < 1².3².5²….199²/[(1.3).(3.5).(5.7)…(199.2...
=1².3².5²….199²/(1.3.3.5.5.7…199.201)
=1².3².5²….199²/(1.3².5².7²…199².201)
=1/201
A=1/2+3/4+5/6+........+199/200
chung to rang A^2<1/200
cho A=1/2.3/4.5/6.7/8.....199/200
chung to rang A2<1/201
cho a=1/2*3/4*5/6*...*199/200.chứng minh rằng A^2<1/201
Cho A = 1/2 + 3/4 + 5/6 + .....+ 199/200
Chứng minh A2 < 1/201
A=\(\frac{1.2+2.3+3.4+.....+20.21}{1+2-3-4+5+6-7-8+......+197+198-199-200+201}\)
Tính A
Tính tông sau một cách hợp lí
A= 1-3+6-7+.....+97-99+101
B= -1-2-3-4-5 -........-199-200-201-202
1 tính
a,( 17/20+19/21). 31/20 (17/20+19/21). (-31/20)
b, 3/4.15/2+15/8.(-3) + 199/200
c, (199/200+201/202+2015/2014). (1/2+2/3+ -7/6)
d, a(b+1/5) -a ( 6/5 + b ) với a= 2015/2015 ; b= 2000
Cho A=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\).CMR \(A^2\)<\(\frac{1}{201}\)
Cho \(A=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{199}{200}\)và chứng minh \(A^2< \frac{1}{201}\)
