a.tính A=2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+….+ 2/2017.2019
b cho S = 1/31+ 1/32+ 1/33….+ 1/60.Chứng minh S bé hơn 4/5
a.tính A=2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+….+ 2/2017.2019
b cho S = 1/31+ 1/32+ 1/33….+ 1/60.Chứng minh S bé hơn 4/5
Giải thích các bước giải:
a, A=2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+….+ 2/2017.2019
= 2. (1/1.3 + 1/3.5 +1/5.7 + 1/7.9+….+ 1/2017.2019 )
= 2. (1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+….+1/2017-1/2019)
= 2. (1-1/2019)
= 2.2018/2019
= $\frac{4036}{2019}$
b, 1/31+ 1/32+ 1/33….+ 1/60
(1/31+ 1/32+ 1/33+…+1/40) + (1/41+ 1/42+ 1/43+…+1/50) + (1/51+ 1/52+ 1/53+…+1/60)
Ta tha’y :
1/31+ 1/32+ 1/33+…+1/40 < 1/3
1/41+ 1/42+ 1/43+…+1/50 < 1/4
1/51+ 1/52+ 1/53+…+1/60 < 1/5
⇒ S = 1/3+1/4+1/5
⇒ 47/60 < $\frac{4}{5}$
Hay 47/60 < $\frac{4}{5}$
Xin cau tra loi hay nhat.
Giải thích các bước giải:
a, A = 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9+….+ 2/2017.2019
A = 1 – 1/3 + 1/3 – 1/5 + 1/5 – 1/7 + … + 1/2017 – 1/2019
A = 1 – 1/2019
A = 2018/2019
b, S = (1/31+1/32+1/33+…+1/40) + (1/41 + 1/42 + …+ 1/50) + (1/51 + 1/52+…+1/59+1/60)
Mà : (1/31+1/32+1/33+…+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + …+ 1/50) < 1/4 ; (1/51 + 1/52+…+1/59+1/60) < 1/5
Mà S = (1/3 + 1/4 + 1/5) < 4/5 (Vì 1/3 + 1/5 < 3/5 hay 7/12 < 3/5 hay 35/60 < 36/60)
Vậy S < 4/5