D = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\)
Những câu hỏi liên quan
D=\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
E=\(\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}....\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}\)
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
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Tính: \(\left(\dfrac{1000}{1}+\dfrac{999}{2}+\dfrac{998}{3}+...+\dfrac{2}{999}+\dfrac{1}{1000}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1001}\right)\)
Tính: \(\left(\dfrac{1000}{1}+\dfrac{999}{2}+\dfrac{998}{3}+...+\dfrac{2}{999}+\dfrac{1}{1000}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1001}\right)\)
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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1. Tính:
B=\(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+....+\dfrac{1000}{2^{1000}}\)
\(B=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{3^2}+.....+\dfrac{1000}{2^{1000}}\)
\(2B=2\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{3^3}+.....+\dfrac{1000}{2^{1000}}\right)\)
\(2B=1+1+\dfrac{3}{2^2}+......+\dfrac{1000}{2^{999}}\)
\(2B-B=\left(2+\dfrac{3}{2^2}+.....+\dfrac{1000}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.....+\dfrac{1000}{2^{999}}\right)\)\(2B-B=2-\dfrac{1}{2}-\dfrac{2}{2^2}-\dfrac{1000}{2^{999}}\)
\(B=1-\dfrac{1000}{2^{999}}\)
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\(A=\dfrac{1001}{1000^2+1}+\dfrac{1001}{1000^2+2}+\dfrac{1001}{1000^3+3}+.....+\dfrac{1001}{1000^2+100}\)Chứng minh rằng 1<A2<4
Tính nhanh : \(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+...+\dfrac{1}{998}.\dfrac{1}{999}+\dfrac{1}{999}.\dfrac{1}{1000}\)
\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+...+\dfrac{1}{999}.\dfrac{1}{1000}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\\ =1-\dfrac{1}{1000}=\dfrac{999}{1000}\)
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ta có
1/1.1/2=1-1/2
1/2.1/3=1/2-1/3
1/3.1/4=1/3-1/4
............
1/999.1/1000=1/999-1/1000
Từ đó suy ra
1/1.1/2+1/2-1/3+1/3+.......+1/998.1/999+1/999.1/1000
=1/1-1/2+1/2-1/3+1/3-.....+1/998-1/999+1/999-1/1000
=1-1/1000
=1000/1000-1/1000
=999/1000
nhớ like bạn nhé
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CHứng minh rằng 1 + \(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^{1999}}>1000\)
Tính A biết \(A=\dfrac{1000}{1}+\dfrac{999}{2}+\dfrac{998}{3}+...+\dfrac{2}{999}+\dfrac{1}{1000}\)
Yêu cầu bài toán chỉ đơn thuần tính cái này thôi à em!
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