10
Xét khai triển:
\(\left(x+2\right)^{2013}=C_{2013}^0.2^{2013}+C_{2013}^1x.2^{2012}+C_{2013}^2x^22^{2011}+...+C_{2013}^{2013}x^{2013}\)
Đạo hàm 2 vế:
\(2013\left(x+2\right)^{2012}=C_{2013}^12^{2012}+2C_{2013}^2x.2^{2011}+...+2013C_{2013}^{2013}x^{2012}\)
\(\Rightarrow2013x\left(x+2\right)^{2012}=xC_{2013}^12^{2012}+2C_{2013}^2x^2.2^{2011}+...+2013C_{2013}^{2013}x^{2013}\)
Tiếp tục đạo hàm 2 vế:
\(2013\left(x+2\right)^{2012}+2012.2013x\left(x+2\right)^{2011}=C_{2013}^12^{2012}+2^2C_{2013}^2x.2^{2011}+...+2013^2C_{2013}^{2013}x^{2012}\)
Thay \(x=1\)
\(\Rightarrow2013.3^{2012}+2012.2013.3^{2011}=1^2C_{2013}^12^{2012}+2^2C_{2013}^22^{2011}+...+2013^2C_{2013}^{2013}\)
\(\Rightarrow S=2013.3^{2012}+2012.2013.3^{2011}=2013.3^{2011}\left(3+2012\right)=2013.2015.3^{2011}\)
11.
Ta có:
\(\dfrac{1}{k+1}C_n^k=\dfrac{1}{k+1}.\dfrac{n!}{k!\left(n-k\right)!}=\dfrac{n!}{\left(k+1\right)!\left(n-k\right)!}=\dfrac{1}{n+1}.\dfrac{\left(n+1\right)!}{\left(k+1\right)!\left[\left(n+1\right)-\left(k+1\right)\right]!}\)
\(=\dfrac{1}{n+1}C_{n+1}^{k+1}\)
Do đó:
\(S=\dfrac{1}{2}.C_{2n}^1+\dfrac{2}{4}C_{2n}^3+...+\dfrac{1}{2n}C_{2n}^{2n-1}\)
\(=\dfrac{1}{2n+1}C_{2n+1}^2+\dfrac{1}{2n+1}C_{2n+1}^4+...+\dfrac{1}{2n+1}C_{2n+1}^{2n}\)
\(=\dfrac{1}{2n+1}\left(C_{2n+1}^2+C_{2n+1}^4+...+C_{2n+1}^{2n}\right)\)
Xét khai triển:
\(\left(1+x\right)^{2n+1}=C_{2n+1}^0+C_{2n+1}^1x+C_{2n+1}^2x^2+...+C_{2n+1}^{2n+1}x^{2n+1}\) (1)
Thay \(x=1\) vào (1):
\(\Rightarrow2^{2n+1}=C_{2n+1}^0+C_{2n+1}^1+C_{2n+1}^2+...+C_{2n+1}^{2n+1}\) (2)
Thay \(x=-1\) vào (1):
\(0=C_{2n+1}^0-C_{2n+1}^1x+C_{2n+1}^2-....+C_{2n+1}^{2n}-C_{2n+1}^{2n+1}\) (3)
Cộng vế (2) và (3):
\(2^{2n+1}=2C_{2n+1}^0+2C_{2n+1}^2+...+2C_{2n+1}^{2n}\)
\(\Rightarrow C_{2n+1}^0+C_{2n+1}^2+C_{2n+1}^4+...+C_{2n+1}^{2n}=2^{2n}\)
\(\Rightarrow C_{2n+1}^2+C_{2n+1}^4+...+C_{2n+1}^{2n}=2^{2n}-C_{2n+1}^0=2^{2n}-1\)
\(\Rightarrow S=\dfrac{1}{2n+1}\left(2^{2n}-1\right)=\dfrac{2^{2n}-1}{2n+1}\)
