Question

A_{n} ^ 2 -C n+1 ^ n-1 =4n+6 .

Answer 1

\(A^2_n-C^{n-1}_{n+1}=4n+6\) \(\left(Đkxđ:n\ge2\right)\)

\(\Leftrightarrow A^2_n-C^2_{n+1}=4n+6\)

\(\Leftrightarrow n\left(n-1\right)-\dfrac{n\left(n+1\right)}{2!}=4n+6\)

\(\Leftrightarrow n^2-n-\dfrac{n^2+n}{2}=4n+6\)

\(\Leftrightarrow2n^2-2n-n^2-n=8n+12\)

\(\Leftrightarrow n^2-11n-12=0\)

\(\Leftrightarrow\left[{}\begin{matrix}n=12\left(TM\right)\\n=-1\left(loại\right)\end{matrix}\right.\)

Vậy n=12

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