1+2+...+1000=?
Những câu hỏi liên quan
Tinh [(1+2012/1)*(1+2012/2)*(1+2012/3)*...*(1+2012/1000)]/[(1+1000/1)*(1+1000/2)*(1+1000/3)*...*(1+1000/2012)]
C ={(1+(1999/1))(1+(1999/2))(1+(1999/3))+...+(1+(1999/1000))}/{(1+(1000/1))(1+(1000/2))(1+(1000/3))...(1+(1000/1999))}
A=(1+1999/1).(1+1992/2).(1+1999/3)...(1+1999/1000)/(1+1000/1).(1+1000/2).(1+1000/3)...(1+1000/1999)
Tính A
A = (1 + 1999/1)(1 + 1999/2)......(1 + 1999/1000)
B = ( 1 + 1000/1)(1 + 1000/2)......(1 + 1000/1999)
Tính A/B
\(A=\left(1+\dfrac{1999}{1}\right)\left(1+\dfrac{1999}{2}\right)...\left(1+\dfrac{1999}{1000}\right)\)
\(=\dfrac{2000}{1}.\dfrac{2001}{2}.\dfrac{2002}{3}...\dfrac{2999}{1000}\)\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}\)
\(B=\left(1+\dfrac{1000}{1}\right)\left(1+\dfrac{1000}{2}\right)...\left(1+\dfrac{1000}{1999}\right)\)
\(=\dfrac{1001}{1}.\dfrac{1002}{2}.\dfrac{1003}{3}...\dfrac{2999}{1999}\) \(=\dfrac{1001.1002.1003...2999}{1.2.3...1999}\)
\(\Rightarrow A:B=\left(\dfrac{2000.2001.2002...2999}{1.2.3...1000}\right):\left(\dfrac{1001.1002.1003...2999}{1.2.3...1999}\right)\)
\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}.\dfrac{1.2.3...1999}{1001.1002.1003...2999}\)
\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}.\dfrac{1.2.3...1000.\left(1001.1002...1999\right)}{1001.1002.1003....1999.\left(2000.2001.2002.2999\right)}\)\(=\dfrac{1.2.3...1000}{1.2.3...1000}=1\)
Vậy \(\dfrac{A}{B}=1\)
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Cho A = \(\dfrac{1001}{1000^2+1}\)+\(\dfrac{1001}{1000^2+2}\)+\(\dfrac{1001}{1000^2+3}\)+...+\(\dfrac{1001}{1000^2+1000}\)
Chứng minh rằng 1<\(^{A^2}\)<4
Tổng A có 1000 số hạng.
Vậy
Chúc bạn học tốt.
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Tổng A có 1000 số hạng
A>(1001/1000^2+1000)*1000=1001*1000/1000*(1000+1)=1
A<(1001/1000^2)*1000=1001/1000=1+1/1000<1
Vậy 1<A<2 nên 1<A^2<4
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tính: B=[(1+2012/1)+(1+2012/2)+....+(1+2012/1000)]:[(1+1000/1)+(1+1000/2)+....+(1+1000/2012)]
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tính B=[(1+2012/1)+(1+2012/2)+(1+2012/3)+...+(1+2012/1000)]:[(1+1000/1)+(1+1000/2)+...+(1+1000/2012)]
101/1000^2+1 + 101/1000^2+2 + ... + 101/1000^2+1000
Cái này là a) \(\frac{101}{1000^2+1}\) hay là b)\(\frac{101}{1000^2}+1\)vậy nhỉ?
cái này em ko biết nha em lớp 4
Tinh
(1 + 1999/1)(1 + 1999/2)......(1 + 1999/1000)
( 1 + 1000/1)(1 + 1000/2)......(1 + 1000/1999)
Tinh
(1 + 1999/1)(1 + 1999/2)......(1 + 1999/1000)
( 1 + 1000/1)(1 + 1000/2)......(1 + 1000/1999)
