1,so sánh: √1994 +√1996 và 2√1995
2,rút gọn :S=√(1+1/(2^2)+1/(3^2)) 3√(1+1/(3^2)+1/(4^2)) +...+√(1+1/(2005^2)+1/(2006^2))
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TÍnh
a,S = 2 - 4 + 6 - 8 +.....+ 1998 - 2000
b,S = 2 - 4 - 6 + 8 + 10 - 12 - 14 + 16 + ...... + 1994 - 1996 - 1998 + 2000
c,- 1 - 2 - 3 - ... - 2005 - 2006 - 2007
a) Số số hạng: (200-2):2+1=100\(\Rightarrow\)S=(2-4)+(6-8)+...+(1998-2000)=-2x50=-100
b) S=(2-4)-(6-8)-...-(1994-1996)-(1998-2000)=0
c) S=-(1+2+3+....+2005+2008+2007)
Số số hạng:(2007-1)+1=2007. Vậy S=-(2007+1)x2007:2=-2015028
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Cho S= \(\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}+1}+........+\frac{2^{n+1}}{2005^{2^n}+1}+.......+\frac{2^{2006}}{2005^{2^{2006}}+1}\)
So sánh S với \(\frac{1}{1002}\)
Cho \(S=\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}+1}+...+\frac{2^{n+1}}{2005^{2^{n+1}}+1}+...+\frac{2^{2006}}{2005^{2^{2006}}+1}\)
So sánh S với \(\frac{1}{1002}\)
E=\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+.....+\sqrt{1+\frac{1}{2005^2}+\frac{1}{2006^2}}\)
Rút gọn E
Ap dung cong thuc \(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=1+\frac{1}{a}-\frac{1}{a+1}\)
ta co \(E=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2005}-\frac{1}{2006}=2004+\frac{1}{2}-\frac{1}{2006}\)
Ta có:
\(E=\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{\left(-3\right)^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{\left(-4\right)^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{2005^2}+\frac{1}{\left(-2006\right)^2}}\)
DO: \(1+2+\left(-3\right)=0;1+3+\left(-4\right)=0;...;1+2005+\left(-2006\right)=0\)
=> TA ĐƯỢC: \(E=\sqrt{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{-3}\right)^2}+\sqrt{\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{-4}\right)^2}+...+\sqrt{\left(\frac{1}{1}+\frac{1}{2005}+\frac{1}{-2006}\right)^2}\)
=> \(E=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{1}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1}+\frac{1}{2005}-\frac{1}{2006}\)
=> \(E=\left(\frac{1}{1}+\frac{1}{1}+...+\frac{1}{1}\right)+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\right)\)
DO TRONG E CÓ TẤT CẢ 2004 CĂN THỨC
=> \(E=2004+\frac{1}{2}-\frac{1}{2006}=2004+\frac{501}{1003}=\frac{2010513}{1003}\)
\(S=\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}+1}+...+\frac{2^{n+1}}{2005^{2^n}+1}+...+\frac{2^{2006}}{2005^{2^{2005}}+1}\)So sánh S với \(\frac{1}{1002}\)
Cho S=\(\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}}+...\)\(..+\frac{2^{n+1}}{2005^{2^n}}+...+\frac{2^{2006}}{2005^{2^{2005}}+1}\)
So sánh S với \(\frac{1}{1002}\)
Cho A =1/1-1/2+1/3-1/4+...+1/2005+1/2006
B=(1/1+1/2+1/2006)-2x(1/2+1/4+...+1/2006)
a)So sánh A và B
b)Chứng minh rằng A=1/1004+1/1005+...1/2006
a,\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{2005}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2006}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2006}\right)\)
\(=B\left(ĐPCM\right)\)
b, \(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2006}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1003}\right)\)
\(A=\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2006}\)
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ui ghi lộn, chữ đpcm chuyển xuống dòng cuối cùng nhé :v
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Cho \(S=\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+...+\frac{2^{n+1}}{2005^{^{2^n}}+1}+...+\frac{2^{2006}}{2006^{2^{2005}}+1}\). So sánh S với \(\frac{1}{1002}\)
1. Rút gọn S=1+2+2 mũ 2+2 mũ 3+2 mũ 4+...+2 mũ 2014
2. So sánh S và D=2 mũ 2015
Chắc đề thế này!
\(S=1+2+2^2+2^3+2^4+...+2^{2014}\)
\(2S=2+2^2+2^3+2^4+...+2^{2015}\)
\(2S-S=\left(2+2^2+2^3+...+2^{2015}\right)-\left(1+2+2^2+...+2^{2014}\right)\)
\(\Rightarrow2S-S=S=2^{2015}-1< 2^{2015}\Rightarrow S< D\)