Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta co:\(B=\frac{2008}{1}+\frac{2007}{2}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{2009-1}{1}+\frac{2009-2}{2}+...+\frac{2009-2007}{2007}+\frac{2009-2008}{2008}\)
\(B=\left(\frac{2009}{1}+\frac{2009}{2}+...+\frac{2009}{2008}\right)-\left(\frac{1}{1}+\frac{2}{2}+...+\frac{2008}{2008}\right)\)
\(B=2009+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)-2008\)
\(B=1+2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right)\)
\(B=2009\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2008}+\frac{1}{2009}\right)\)
Vay \(\frac{A}{B}=\frac{1}{2009}\)
Nhân vô rồi chuyển dấu lên và nhóm nhân -1ra ngoài rồi trg ngoặc là dãy có quy luật giải dãy đó r nhân phá ngoặc
\(a_{n-1}=\frac{1}{1+2+3+...+n}=\frac{2}{n\left(n+1\right)}\)=>\(1-a_{n-1}=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
\(A=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)........\left(1-\frac{2}{2006.2007}\right)\)
\(=\left(\frac{1.4}{2.3}\right)\left(\frac{2.5}{3.4}\right)\left(\frac{3.6}{4.5}\right)........\left(\frac{2005.2008}{2006.2007}\right)\)\(=\frac{\left(1.2.3......2005\right)\left(4.5.6.....2008\right)}{\left(2.3.4.....2006\right)\left(3.4.5....2007\right)}=\frac{1.2008}{2006.3}=\frac{1004}{3009}\)
Vế trái:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
=\(\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)=Vế phải
\(\frac{1}{1-\frac{1}{1-\frac{1}{2}}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}\)
\(=\frac{1}{1-\frac{1}{\frac{1}{2}}}+\frac{1}{1+\frac{1}{\frac{3}{2}}}\)
\(=\frac{1}{1-2}+\frac{1}{1+\frac{2}{3}}=-1+\frac{1}{\frac{5}{3}}\)
\(=-1+\frac{3}{5}=\frac{-2}{5}\)
\(\frac{1}{1-\frac{1}{1-\frac{1}{2}}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}} =\frac{1}{\frac{1-\frac{1}{2}-1}{1-\frac{1}{2}}}+\frac{1}{\frac{1+\frac{1}{2}+1}{1+\frac{1}{2}}}\)
\(=\frac{1-\frac{1}{2}}{\frac{-1}{2}}+\frac{1+\frac{1}{2}}{2+\frac{1}{2}}=\frac{\frac{2-1}{2}}{\frac{-1}{2}}+\frac{\frac{2+1}{2}}{\frac{4+1}{2}}\)
\(=\frac{\frac{3}{2}}{\frac{-1}{2}}+\frac{\frac{3}{2}}{\frac{5}{2}}=\frac{3}{2}.\frac{2}{-1}+\frac{3}{2}.\frac{2}{5}\)
\(=-3+\frac{3}{5}=\frac{-15+3}{5}=\frac{-12}{5}\)