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 \(S=\dfrac{2^{2018}-1}{2018}\)

999 - 888 - 111 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111 + 111 - 111

= 0 + 111 - 111 + 111 - 111

= 111 - 111 + 111 - 111

= 0 + 111 - 111

= 111 - 111

= 0

\(\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\left(\frac{2018}{2017}+\frac{2017}{2016}+\frac{2016}{2017}\right)\)

= \(\left(\frac{3-2-1}{6}\right)\left(\frac{2018}{2017}+\frac{2017}{2016}+\frac{2016}{2017}\right)\)

\(=0\cdot\left(\frac{2018}{2017}+\frac{2017}{2016}+\frac{2016}{2017}\right)=0\)

Bài đây dễ mà :vv

Ta có: \(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)

\(=1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)\)

\(=\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2018}\)

\(=2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)

Giờ ta thế vào bài toán ban đầu được

\(A=\frac{\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)

\(=\frac{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}\)

\(=\frac{2017}{2018}\)  

Bằng \(-\frac{2}{11}\)

D. Tìm x thuộc Z biết 

x+(x+1)+(x+2)+....+2016+2017=2017 

=> ( x + x + x + ..+ x ) + ( 1 + 2 + 3+...+2016 + 2017 ) = 2017 

<=> 2017x + 2035153 = 2017 

=> 2017x = -2033136

=> x = -1008

Vậy ...

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