2000<20001
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\(\frac{2000}{1.2}+...+\frac{2000}{2002.2003}\)
\(=2000.\left(\frac{1}{1.2}+....+\frac{1}{2002.2003}\right)\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2002}-\frac{1}{2003}\right)
\)
\(=2000.\left(\frac{1}{1}-\frac{1}{2003}\right)=2000.\frac{2002}{2003}\)
đặt A=200/1.2+200/2.3+200/3.4+...+200/2002.2003
A:2000 = 1-1/2+1/2-1/3+...+1/2002-1/2003
A:2000=1-1/2003
A:2000=2002/2003
A=....
k nhe
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2000.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=2000.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
\(=2000.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)=2000.\left(1-\frac{1}{100}\right)=20.99=1980\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét khai triển:
\(\left(x+1\right)^n=C_n^0+C_n^1x+C_n^2x^2+...+C_n^nx^n\)
\(\Leftrightarrow x\left(x+1\right)^n=C_n^0.x+C_n^1x^2+C_n^2x^3+...+C_n^nx^{n+1}\)
Thay \(n=2000\) ta được:
\(x\left(x+1\right)^{2000}=C_{2000}^0x+C_{2000}^1x^2+C_{2000}^2x^3+...+C_{2000}^{2000}x^{2001}\)
Đạo hàm 2 vế:
\(\left(x+1\right)^{2000}+2000x\left(x+1\right)^{1999}=C_{2000}^0+2C_{2000}^1x+...+2001C_{2000}^{2000}x^{2000}\)
Thay \(x=1\) ta được:
\(2^{2000}+2000.2^{1999}=C_{2000}^0+2C_{2000}^1+...+2001.C_{2000}^{2000}\)
\(\Rightarrow S=2^{1999}\left(2+2000\right)=2002.2^{1999}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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