Câu 20.
\(C_n^2+C_n^3=4n\)
Đk: \(n\ge3\)
Pt\(\Rightarrow\dfrac{n!}{2!\left(n-2\right)!}+\dfrac{n!}{3!\left(n-3\right)!}=4n\)
\(\Rightarrow\dfrac{n\left(n-1\right)\left(n-2\right)!}{2\left(n-2\right)!}+\dfrac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{6\left(n-3\right)!}=4n\)
\(\Rightarrow\dfrac{n\left(n-1\right)}{2}+\dfrac{n\left(n-1\right)\left(n-2\right)}{6}=4n\)
Chia cả hai vế cho \(n\) ta được:
\(\Rightarrow\dfrac{n-1}{2}+\dfrac{\left(n-1\right)\left(n-2\right)}{6}=4\)
Bạn tự quy đồng giải pt bậc hai tìm n nhé.
Câu 21:
\(\frac{1}{2}A^2_{2x}-A^2_x\leq \frac{6}{x}C^3_x+10\)
\(\Leftrightarrow \frac{1}{2}.\frac{(2x)!}{(2x-2)!}-\frac{x!}{(x-2)!}\leq \frac{6}{x}.\frac{x!}{3!(x-3)!}+10\)
\(\Leftrightarrow \frac{1}{2}.2x(2x-1)-(x-1)x\leq (x-1)(x-2)+10\)
\(\Leftrightarrow 12-3x\geq 0\Leftrightarrow x\leq 4\)
Mà $x$ tự nhiên, $x\geq 3$ nên $x=3, x=4$
Đáp án C.
Câu 20:
\(C^2_n+C^3_n=4n\)
\(\Leftrightarrow\dfrac{n.\left(n-1\right)}{2!}+\dfrac{4.\left(n-1\right).\left(n-2\right)}{3!}=4n\)
\(\Leftrightarrow\dfrac{n.\left(n-1\right)}{2}+\dfrac{n.\left(n-1\right).\left(n-2\right)}{6}=4n\)
\(\Leftrightarrow3n.\left(n-1\right)+n.\left(n-1\right).\left(n-2\right)=24n\)
\(\Leftrightarrow3n^2-3n+n.\left(n^2-3n+2\right)-24n=0\)
\(\Leftrightarrow n^3-25n=0\)
\(\Leftrightarrow n.\left(n^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=0\\n^2=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=0\\\left[{}\begin{matrix}n=5\\n=-5\end{matrix}\right.\end{matrix}\right.\)
