Question

If ${}^{2n+3}{C}_{2n}-{}^{2n+2}{C}_{2n-1}=15\left(2n+1\right)$, then $n=$

• A. $13$
• B. $14$
• C. $27$
• D. $15$

Answer 1

Answer: (B)

$\frac{\lfloor 2n+3}{\lfloor 2n\lfloor 3}-\frac{\lfloor 2n+2}{\lfloor 2n-1\lfloor 3}=15\left(2n+1\right)$
$\Rightarrow \frac{\left(2n+2\right)\left(2n+1\right)}{2}=15\left(2n+1\right)$
$\Rightarrow n+1=15$
$\Rightarrow n=14.$