tính tổng \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
Những câu hỏi liên quan
Tính Q =\(\dfrac{1.3}{3.5}+\dfrac{2.4}{5.7}+\dfrac{3.5}{7.9}+...+\dfrac{\left(n-1\right)\left(n+1\right)}{\left(2n-1\right)\left(2n+1\right)}+...+\dfrac{1002.1004}{2005.2007}\)
Cho \(S_n=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\). Kết quả giới hạn của limSn bằng:
A. \(\dfrac{1}{3}\)
B. \(\dfrac{1}{5}\)
C. \(\dfrac{1}{2}\)
D. 0
\(S_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2n}{2n+1}\right)=\dfrac{2n}{2\left(2n+1\right)}\)
-> chọn D
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Xem thêm câu trả lời
Cho dãy số ($u_n$) biết \(u_{n+1}=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\) . Xác dịnh số hạng thứ 50 của dãy số
\(u_{n+1}=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(2n-1\right)\cdot\left(2n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{n}{2n+1}\)
=>\(u_{50}=u_{49+1}=\dfrac{49}{2\cdot49+1}=\dfrac{49}{99}\)
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Chứng minh rằng:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}=\dfrac{n+1}{2n+3}\)
Giúp mik vs các bn
\(P=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\\ 2P=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2n+1\right)\left(2n+3\right)}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\\ =1-\dfrac{1}{2n+3}\\ =\dfrac{2\left(n+1\right)}{2n+3}\\ P=\dfrac{2\left(n+1\right)}{2n+3}:2\\ =\dfrac{n+1}{2n+3}\)
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Tính các tổng sau :
a) \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
b) \(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.11}+...+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)
c) \(\dfrac{7}{1.8}+\dfrac{7}{8.15}+\dfrac{7}{15.22}+...+\dfrac{1}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}=\dfrac{1}{2}\cdot\dfrac{2n}{2n+1}=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
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Tính các tổng sau :
a, \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+......+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
b, \(\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.11}+........+\dfrac{1}{\left(4n-3\right)\left(4n+1\right)}\)
c,\(\dfrac{7}{1.8}+\dfrac{7}{8.15}+\dfrac{7}{15.22}+....+\dfrac{1}{\left(7n-6\right)\left(7n+1\right)}+\dfrac{1}{7n+1}\)
a: \(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2n+1-1}{2n+1}\)
\(=\dfrac{n}{2n+1}\)
b: \(=\dfrac{1}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{\left(4n-3\right)\left(4n+1\right)}\right)\)
\(=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{4n-3}-\dfrac{1}{4n+1}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{4n}{4n+1}=\dfrac{n}{4n+1}\)
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Tính
lim \(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+.....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\lim\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\right)\)
\(=\lim\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\)
\(=\lim\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\)
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Tìm tổng :
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right).\left(2n+1\right)}+...+\dfrac{1}{255.257}\)
Ta có:\(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}=\dfrac{2}{\left(2n-1\right).\left(2n+1\right)}\)
Ta phân tích tổng thành:
\(\dfrac{1}{2}.\left[\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(2n-1\right).\left(2n+1\right)}+...+\dfrac{2}{255.257}\right]\)
\(=\dfrac{1}{2}.\left[\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{255}-\dfrac{1}{257}\right]\)
\(=\dfrac{1}{2}.\left[1-\dfrac{1}{257}\right]=\dfrac{128}{257}\)
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Chứng minh BĐT sau
a)\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}< \dfrac{1}{2}\)
b)
a)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}\)
P/s: Cj chỉ biết làm ý a thôi nhé! Có j ko hiểu cmt nhé!
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