2+2.4
3/2.4^x+5/3.4^x 2=3/2.4^8+5/3.4^10
Biết \(lim\dfrac{\sqrt{2.4^n+1}-2^n}{\sqrt{2.4^n+1}+2^n}=a+b\sqrt{2}\left(a,b\in Z\right)\). Tính \(a^3+b^3\)
\(\lim\limits\dfrac{\sqrt{2\cdot4^n+1}-2^n}{\sqrt{2\cdot4^n+1}+2^n}\)
\(=\lim\limits\dfrac{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}-2^n}{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}+2^n}\)
\(=\lim\limits\dfrac{\sqrt{2+\dfrac{1}{4^n}}-1}{\sqrt{2+\dfrac{1}{4^n}}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-2\sqrt{2}}{2-1}=3-2\sqrt{2}\)
=>a=3; b=-2
\(a^3+b^3=3^3+\left(-2\right)^3=27-8=19\)
Chứng minh rằng:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.4^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+....+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+....+\frac{1}{9^2}-\frac{1}{10^2}=\frac{1}{1^2}-\frac{1}{10^2}<1\)
=>đpcm
tinh tổng :
a. S = 0.2 + 2.4 + 4.6 +.......+ 98.100
b. S= 0.2+ 2.4 + 4.6 + .........+ 2n. ( 2n +2 )
Tinh:
a/S=0.2+2.4+4.6+......+98.100
b/ A= 0.2+2.4+4.6+......+ 2n(2n+2)
2/2.4+2/4.6+2/6.8+...+2/98.100
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{98.100}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{98}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}\\ =\dfrac{49}{100}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{98.100}\)\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+....+\dfrac{1}{98}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
2+2.4 = ?
2/2.4+2/4.6+....+2/50.52
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+................+\dfrac{2}{50.52}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..............+\dfrac{1}{50}-\dfrac{1}{52}\)
\(=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+.....+\dfrac{2}{50.52}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+.....+\dfrac{1}{50}-\dfrac{1}{52}\)
\(=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
2/2.4+2/4.6+....+2/2010.2012
Đặt A = 2/2.4+2/4.6+....+2/2010.2012
= 1/2 - 1/4 + 1/4 - 1/6 + ...... + 1/2010 - 1/2012
= 1/2 - 1/2012
= 1005/2012
2/2.4+2/4.6+...+2/2010.2012
= 1/2-1/4+1/4-1/6+...+1/2010-1/2012
= 1/2-1/2012
= 1006/2012-1/2012
= 1005/2012
1/2-1/4+1/4-1/6+....+1/2010-1/2012
2
1/2-1/2012
2
1005/2024