Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\dfrac{1}{2018!\left(2019-2018\right)!}+\dfrac{1}{2016!\left(2019-2016\right)!}+...+\dfrac{1}{2!\left(2019-2\right)!}+\dfrac{1}{0!\left(2019-0!\right)}\)
\(\Rightarrow2019!.S=\dfrac{2019!}{2018!\left(2019-2018\right)!}+\dfrac{2019!}{2016!\left(2019-2016\right)!}+...+\dfrac{2019!}{2!\left(2019-2\right)!}+\dfrac{2019!}{0!\left(2019-0\right)!}\)
\(=C_{2019}^{2018}+C_{2019}^{2016}+...+C_{2019}^2+C_{2019}^0\)
\(=\dfrac{1}{2}\left(C_{2019}^0+C_{2019}^1+...+C_{2019}^{2018}+C_{2019}^{2019}\right)\)
\(=\dfrac{1}{2}.2^{2019}=2^{2018}\)
\(\Rightarrow S=\dfrac{2^{2018}}{2019!}\)
Xét khai triển:
\(\left(1+x\right)^{2017}=C_{2017}^0+xC_{2017}^1+x^2C_{2017}^2+...+x^{2017}C_{2017}^{2017}\)
Lấy tích phân 2 vế:
\(\int\limits^1_0\left(1+x\right)^{2017}=\int\limits^1_0\left(C_{2017}^0+xC_{2017}^1+...+x^{2017}C_{2017}^{2017}\right)\)
\(\Leftrightarrow\dfrac{2^{2018}-1}{2018}=C_{2017}^0+\dfrac{1}{2}C_{2017}^1+...+\dfrac{1}{2018}C_{2017}^{2017}\)
Vậy \(S=\dfrac{2^{2018}-1}{2018}\)
Xét 2 khai triển:
\(\left(x+1\right)^{2018}=C_{2018}^0+C_{2018}^1x+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\)
\(\left(x-1\right)^{2018}=C_{2018}^0-C_{2018}^1x+C_{2018}^2x^2-...+C_{2018}^{2018}x^{2018}\)
Cộng vế với vế:
\(\left(x+1\right)^{2018}+\left(x-1\right)^{2018}=2\left(C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\right)\)
\(\Leftrightarrow C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}=\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}=\frac{\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}-2^{2017}}{x-1}=\lim\limits_{x\rightarrow1}\frac{1009\left(x+1\right)^{2017}+1009\left(x-1\right)^{2017}}{1}=1009.2^{2017}\)
Xét số hạng tổng quát.
Cho k chạy từ 1 đến 2018 ta được:
Cho x=1 suy ra 
Chọn A.
