LD
1
K
Khách
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S
1
LH
10 tháng 5 2021
\(M=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{8}+...+\frac{197}{198}-\frac{199}{200}\)
\(=\left(1-\frac{1}{2}\right)-\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{6}\right)-\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{198}\right)-\left(1-\frac{1}{200}\right)\)=\(=-\frac{1}{2}+\frac{1}{4}-\frac{1}{6}+\frac{1}{8}-...-\frac{1}{198}+\frac{1}{200}\)
\(=-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=-\frac{1}{2}\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\right]\)
\(=-\frac{1}{2}\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\)
\(=-\frac{1}{2}\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)
\(=-\frac{1}{2}.N\)
\(Tacó:\)
\(M:N=-\frac{1}{2}.N:N=-\frac{1}{2}\)
H
0
NN
0
PT
20 tháng 3 2017
Ta có:
\(1.2.3....99=\dfrac{1.2.3.4....99.100}{2.4.6....100}\)
\(=\dfrac{1}{2.1}.\dfrac{2}{2.2}...\dfrac{100}{2.50}\)
\(=\dfrac{1.2.3.4....100}{1.2.3....50.2.2.2...2}\) ( \(50\) thừa số \(2\) )
\(=\dfrac{51.51...100}{2.2.2...2}\)
\(=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)
\(\Rightarrow A=B\left(đpcm\right)\)
NT
0
NT
1
24 tháng 4 2017
Ta có:
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}=\dfrac{51.52.53...100}{2^{50}}\)
\(=\dfrac{\left(51.52.53..100\right)\left(1.2.3.4...50\right)}{2^{50}\left(1.2.3.4...50\right)}\)
\(=\dfrac{1.2.3.4.5.6...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)
\(=\dfrac{1.2.3.4.5.6...100}{2.4.6.8.10...100}=\dfrac{\left(1.3...99\right)\left(2.4...100\right)}{2.4.6...100}\)
\(=1.3.5.7.99=A\)
Vậy \(A=B\) (Đpcm)
\(M=1+5+5^2+...+5^{2005}\)
\(\Rightarrow5M=5+5^2+5^3+...+5+5^{2006}\)
\(\Rightarrow5M-M=\left(5+5^2+...+5^{2006}\right)-\left(1+5+...+5^{2005}\right)\)
\(\Rightarrow5M-M=4M=5^{2006}-1\Rightarrow M=\frac{5^{2006}-1}{4}\)
\(\frac{N}{4}=\frac{5^{2006}}{4}>\frac{5^{2006}-1}{4}=M\Rightarrow M< \frac{N}{4}\)