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Mathematics Question on permutations and combinations

If 2n+3C2n− 2n+2C2n−1=15(2n+1) ^{2n+3}C_{2n} - \,^{2n+2}C_{2n-1} = 15 \left(2n+1\right), then n=n=

A

1313

B

1414

C

2727

D

1515

Answer

1414

Explanation

Solution

⌊2n+3⌊2n⌊3−⌊2n+2⌊2n−1⌊3=15(2n+1)\frac{\lfloor2n+3 }{\lfloor2n \lfloor3} -\frac{ \lfloor2n+2}{\lfloor2n-1\lfloor3} = 15\left(2n+1\right) ⇒(2n+2)(2n+1)2=15(2n+1)\Rightarrow \frac{\left(2n+2\right)\left(2n+1\right)}{2} = 15\left(2n+1\right) ⇒n+1=15 \Rightarrow n+1=15 ⇒n=14.\Rightarrow n=14.