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Nhận thấy \(\)\(\dfrac{1}{1.1!}=1\); \(\dfrac{1}{2.2!}=\dfrac{1}{4}\)
Đặt \(P=\dfrac{1}{3.3!}+...+\dfrac{1}{2013.2013!}\)
\(P=\dfrac{1}{3.1.2.3}+...+\dfrac{1}{2013.1.2...2013}\)
\(P< \dfrac{1}{1.2.3}+...+\dfrac{1}{2011.2012.2013}\)
\(P< \dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}-\dfrac{1}{2012.2013}\right)\)
\(P< \dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2012.2013}\right)=\dfrac{1}{4}-\dfrac{1}{2.2012.2013}\)
\(P< \dfrac{1}{4}\)
\(A< \dfrac{1}{4}+\dfrac{1}{4}+1=\dfrac{3}{2}\left(đpcm\right)\)
Ta có:
\(\frac{1}{2.2}\)<\(\frac{1}{1.2}\)
\(\frac{1}{3.3}\)<\(\frac{1}{2.3}\)
..............
\(\frac{1}{1009.1009}\)<\(\frac{1}{1008.1009}\)
=>A< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1008.1009}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1008}-\frac{1}{1009}\)
=\(\frac{1}{1}-\frac{1}{1009}=\frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)
=>A<\(\frac{3}{4}\)
Mình nghĩ bạn cần xem lại :
\(A< \frac{1008}{1009}>\frac{1008}{1344}=\frac{3}{4}\)không có nghĩa là \(A< \frac{3}{4}\)
Xem lại ..
1. a) 2B = 1 + 1/2 + 1/22+...+1/298
B - B = (1+1/2+...+1/298) - (1/2+....+1/299)
B = 1 - 299 => B < 1
b) Làm tương tự như câu a, ra là (1 - 1/399) : 2 = 1/2 - 1/2.399(C bé hơh 1/2)
1. a). Theo đầu bài ta có:
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\Leftrightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{99}}< 1\)( đpcm )
1/2^2=4
1/3^2<1/2.3
.................
1/100^2<1/99.100
A<1/4+1/2.3+...+1/99.100
A<1/4+1/2-1/100
A<1/4<3/4
Vậy A<3/4(dpcm).CHÚC BẠN HỌC TỐT!
a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
b) b = a - c => b + c = a
\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)
Bước 2 bạn sai rồi. Vd: \(\frac{1}{3x3}\) đâu bằng hay nhỏ hơn \(\frac{1}{2x3}\)
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~