Tính tổng: S = 1 2018 C 2018 1 2 + 2 2017 C 2018 2 2 + ... + 2017 2 C 2018 2017 2 + 2018 1 C 2018 2018 2
A. S = 1 2018 C 4036 2018
B. S = 1 2018 C 4036 2018
C. S = 2018 2019 C 2018 1009
D. S = 2018 2019 C 4036 2018
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Ta có: \(\left(2018+2017\right)^2>2018^2+2017^2\)
Ta có: \(C=\frac{2018^2-2017^2}{2018^2+2017^2}\)
\(=\frac{\left(2018-2017\right)\left(2018+2017\right)}{2018^2+2017^2}=\frac{2018+2017}{2018^2+2017^2}\)
Ta có: \(D=\frac{2018-2017}{2018+2017}\)
\(=\frac{\left(2018-2017\right)\left(2018+2017\right)}{\left(2018+2017\right)^2}=\frac{2018+2017}{\left(2018+2017\right)^2}\)
Đặt a=2018
b=2017
Ta có: \(\left(2018+2017\right)^2=\left(a+b\right)^2\)
\(2018^2+2017^2=a^2+b^2\)
mà \(\left(2018+2017\right)^2>2018^2+2017^2\)(cmt)
nên \(\left(a+b\right)^2>a^2+b^2\)
\(\Leftrightarrow\frac{a+b}{\left(a+b\right)^2}< \frac{a+b}{a^2+b^2}\)
hay \(\frac{2018+2017}{\left(2018+2017\right)^2}< \frac{2018+2017}{2018^2+2017^2}\)
hay D<C
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
Đáp án là D
Ta có:
S = 1 - 2 + 3 - 4 + ... + 2017 - 2018
S = (1 - 2) + (3 - 4) + ... + (2017 - 2018)
S = (-1) + (-1) + ... + (-1)
S = 1009.(-1) = -1009
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:
\(\left(x+1\right)^{2018}+\left(x-1\right)^{2018}=2\left(C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}\right)\)
\(\Leftrightarrow C_{2018}^0+C_{2018}^2x^2+...+C_{2018}^{2018}x^{2018}=\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}=\frac{\frac{1}{2}\left(x+1\right)^{2018}+\frac{1}{2}\left(x-1\right)^{2018}-2^{2017}}{x-1}=\lim\limits_{x\rightarrow1}\frac{1009\left(x+1\right)^{2017}+1009\left(x-1\right)^{2017}}{1}=1009.2^{2017}\)
Đáp án D
Ta có:
k n C n k 2 = k n n ! k ! n − k ! 2 = C n k . n − 1 ! k − 1 ! n − k ! = C n k . C n − 1 k − 1
Do đó: C 2018 0 . C 2018 1 + C 2018 1 . C 2018 2 + ... + C 2018 2017 . C 2018 2018
Xét khai triển: 1 + x 2018 . x + 1 = 1 + x 4036
Hệ số chứa x 2017 trong khai triển 1 + x 2018 . x + 1 là:
C 2018 0 . C 2018 1 + C 2018 1 . C 2018 2 + ... + C 2018 2017 . C 2018 2018 = S
Hệ số chứa x 2017 trong khai triển 1 + x 4036 là:
C 4036 2017 = 4036 ! 2017 ! .2019 ! = 4036 ! 2018 ! .2018 ! . 2018 2019 = 2018 2019 C 4036 2018
Vậy S = 2018 2019 C 4036 2018