\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}............+\frac{1}{999+1000}+1\)1
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B=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+......................+\frac{4}{999=1000}\)
\(\text{Đề phải như này bạn nha : }B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)
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N Lam theo đề Nguyễn Thiều Công Thành nha :
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{999}-\frac{1}{1000}\)
\(\Rightarrow B=1-\frac{1}{1000}=\frac{999}{1000}\)
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\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{999^2}{999\cdot1000}\)
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{999^2}{999.1000}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.....\frac{999.999}{999.1000}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{999}{1000}\)
\(=\frac{1}{1000}\)
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Tính nhanh:
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.....+\frac{1}{1000\cdot1001}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1000\cdot1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}\)
\(=\frac{1000}{1001}\)
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1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.............+\(\frac{1}{1000}\)-\(\frac{1}{1001}\)=1-\(\frac{1}{1001}\)=\(\frac{1000}{1001}\)
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Xem thêm câu trả lời
Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+...................+\(\frac{1}{999\cdot1000}\)+1
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)
\(=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{1000-999}{999\cdot1000}+1\)
\(=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+...+\frac{1000}{999\cdot1000}-\frac{999}{999\cdot1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{999}{1000}+1\)
\(=\frac{1999}{1000}\)
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\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{1}{k}\times\left(\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\right)\)
Tìm giá trị của k.
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
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\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+.....+\frac{2016}{501}}{\frac{-1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-.....-\frac{1}{999\cdot1000}}\)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+\frac{2016}{998}+...+\frac{2016}{501}}{-\frac{1}{1\cdot2}-\frac{1}{3\cdot4}-\frac{1}{5\cdot6}-...-\frac{1}{999\cdot1000}}\)
\(B=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{999\cdot1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{999}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1000}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{500}\right)}\)
\(B=\frac{2016\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+...+\frac{1}{1000}\right)}\)
\(B=\frac{2016}{-1}=-2016\)
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Tính tổng :a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
Tính tổng :
a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}
b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}
c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}
d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}
e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
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