Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De determinant $\begin{vmatrix}1 & 1 & 1 \\C_n^1 & C_{n+1}^{\: 1} & C_{n+2}^{\: 1} \\ C_n^2 & C_{n+1}^{\: 2} & C_{n+2}^{\: 2} \\\end{vmatrix}$ is gelijk aan	A. 0
	B.
	C. 1
	D. 2
	E. n

[ 5-8679 - op net sinds 6.2.2021-(E)-14.7.2024 ]

Translation in E N G L I S H

The determinant $\begin{vmatrix}1 & 1 & 1 \\C_n^1 & C_{n+1}^{\: 1} & C_{n+2}^{\: 1} \\ C_n^2 & C_{n+1}^{\: 2} & C_{n+2}^{\: 2} \\\end{vmatrix}$ is equal to	A. 0
	B.
	C. 1
	D. 2
	E. n

C_n^p = ⁿC_k = (_kⁿ)

Oplossing - Solution

$\\\cdots=\frac{1}{2}\begin{vmatrix}1&1&1\\n&n+1&n+2\\ n(n-1)&(n+1)n&(n+2)(n+1)\end{vmatrix}=\frac{1}{2}\begin{vmatrix}1&1&1\\ n&n+1&n+2\\n^2-n&n^2+n&n^2+3n+2\end{vmatrix}\newline\\ _{K2\rightarrow K2-K1}^{K3\rightarrow K3-K1}\textrm{=}\frac{1}{2}\begin{vmatrix}1&0&0\\n&1&2\\n^2-n&2n&4n+2\end{vmatrix}=\frac{1}{2}\begin{vmatrix}1&2\\2n&4n+2\end{vmatrix}=\frac{1}{2}(4n+2-4n)=\cdots$