(1-1/2)(1-1/3)....(1-1/199)(1-1/200)
Những câu hỏi liên quan
cmr : -1/2 + 1/3 + -1/4 + ..... + 1/199 + -1/200 = 1/101 + 1/102 + .... + 199 + 1/200
cho A = 1/199+2/198+3197+...+198/2+199/1.Chứng minh A = 200.(1/2+1/3+...+1/200)
\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{189}{2}+\frac{199}{1}\)
\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+199\)
\(A=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
\(A=\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1\)
\(A=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}\)
\(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
Vậy \(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
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CM B=1-1/2+1/3-1/4+.....+1/199-1/200 = 1/101+1/102+.....+1/199+1/200
(x-20)/1*(1/2+1/3+...+1/200)/(1/199+1/998+...+199/1
Tính tỉ số A/B biết:
A= 1/1*2+1/3*4+1/5*6+...+1/199+200
B= 1/101*200+1/102*199+...+1/200*101
Tính tỉ số A/B biết:
A= 1/1*2+1/3*4+1/5*6+...+1/199+200
B= 1/101*200+1/102*199+...+1/200*101
A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
Lại có B = \(\frac{1}{101.200}+\frac{1}{102.199}+...+\frac{1}{200.101}\)
=> 301B = \(\frac{301}{101.200}+\frac{301}{102.199}+...+\frac{301}{200.101}\)
=> 301B = \(\frac{1}{101}+\frac{1}{200}+\frac{1}{102}+\frac{1}{199}+...+\frac{1}{200}+\frac{1}{101}=2\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
=> B = \(\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}{\frac{2}{301}\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)}=\frac{1}{\frac{2}{301}}=\frac{301}{2}=150,5\)
E=1/2+1/3+1/4+...+1/200
199/1+198/2+197/3+...+1/199
S = 1 . 200 + 2 . 199 + 3 . 198 + 4 . 197 + ........ + 199 . 2 + 200 .1
=1*200+2*(200-1)+3*(200-2)+...+199(200-198)+200(200-199)
=(1+2+3+...+200)-(1*2+2*3+...+199*200)
=200*201/2-199*200*201/3
=1353400
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Cho A= ½ + 1/3 + 2/4 +…+ 1/200 và B= 1/199+2/198+3/197+…+199/1
Ta có :
B = 1/ 199 + 2/ 198 + 3/197+...+ 1+ 1 + 1 + ....+ 1. ( tách 199/1 = tổng của 199 số 1)
B = 1 + ( 1+ 1/199) + (1 + 1/198) + ( 1+ 1/197) +....+ (1 + 198/2)
B = 200/200 + 200/199 + 200/198 + 200/197 +...+ 200/2
B = 200 x ( 1/200 + 1/199 + 1/198 + 1/197 +...+ 1/2)
=> A/B =1/ 200
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tinh r=1/2+1/3+1/4+...+1/200:1/199+2/198+3/197+...+198/2+199/1