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B= \(\frac{1}{199}\) + \(\frac{2}{198}\) + ... + \(\frac{198}{2}\) + \(\frac{199}{1}\)
B= ( \(\frac{1}{199}\) + 1) + ( \(\frac{2}{198}\) +1) +...+ ( \(\frac{198}{2}\) +1) +1 ( Mình tách 199 ra thành 199 số hạng rồi cộng thêm vào mỗi phân số)
B= \(\frac{200}{199}\) + \(\frac{200}{198}\) + \(\frac{200}{197}\) +...+\(\frac{200}{2}\)
B= 200( \(\frac{1}{199}\) + \(\frac{1}{198}\) +...+ \(\frac{1}{2}\) )
B= 200 ( \(\frac{1}{2}\) + \(\frac{1}{3}\) +...+ \(\frac{1}{198}\) + \(\frac{1}{199}\) ) = 200 A
Ta thấy A=1A, B=200A Suy ra \(\frac{A}{B}\) = \(\frac{1}{200}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{\frac{(1+2)\cdot2}{2}}+\frac{1}{\frac{(1+3)\cdot3}{2}}+\frac{1}{\frac{(1+4)\cdot4}{2}}+...+\frac{1}{\frac{(1+2012)\cdot2012}{2}}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2025078}}\)
\(Q=\frac{2\cdot2010}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}...+\frac{2}{4050156}}\)
\(Q=\frac{2\cdot2010}{1+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{2012\cdot2013}}\)
\(Q=\frac{2\cdot2010}{1+2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right]}\)
\(Q=\frac{2\cdot2010}{1+2\left[\frac{1}{2}-\frac{1}{2013}\right]}=\frac{2\cdot2010}{1+\frac{2011}{2013}}=\frac{2\cdot2010}{\frac{4024}{2013}}=\frac{4020}{\frac{4024}{2013}}=4020\cdot\frac{2013}{4024}=...\)
Nguyễn Linh Chi ơi , hình như cô nhầm thì phải :v \(2-\frac{2}{2013}=\frac{2\cdot2013-2}{2013}=\frac{4026-2}{2013}=\frac{4024}{2013}\)
sao mà bằng \(\frac{4020}{2013}\)được cô
Ta có:
\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\)
\(P=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)
\(P=1+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{2013.2012}\)
\(P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)\
\(P=1+2\left(\frac{1}{2}-\frac{1}{2013}\right)\)
\(P=1+1-\frac{2}{2013}=2-\frac{2}{2013}=\frac{4020}{2013}\)
\(Q=\frac{2.2010}{P}=\frac{4020}{\frac{4020}{2013}}=2013\)....
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
Bạn xem lời giải của mình nhé:
Giải:
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}+\frac{1}{7.8}\\\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8} \\ =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}< 1\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< 1\)
Chúc bạn học tốt!
Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\)
= \(1-\frac{1}{8}< 1\)
Vậy B < 1
đặt \(A=\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)
\(A=\left(\frac{2003}{2}+1\right)+\left(\frac{2002}{3}+1\right)+..+\left(\frac{1}{2004}+1\right)+\frac{2005}{2005}\)
\(A=\frac{2005}{2}+\frac{2005}{3}+..+\frac{2005}{2004}+\frac{2005}{2005}\)
\(A=2005.\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2004}+\frac{1}{2005}\right)\)
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{A}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2005}}{2005.\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2005}\right)}=\frac{1}{2005}\)
vậy P=1/2005
=1/2 . 2/3 ....1999/2000
=1.2....1999/2.3...2000
1/2000
B= 3/2.4/3. ....2001/2000
B = 3.4....2001/2.3....2000
B =2001/2