In the expansion of (√32^+)n,n N, if the ratio of 15^ term from the beginning to the 15^ t…

Question

In the expansion of ${(\sqrt32}^{+})n,n\in N$, if the ratio of ${15}^{}$ term from the beginning to the ${15}^{}$ term from the end is $\frac{1}{6}$, then the value of ${}^{n}{C}_{3}$ is

Accepted Answer

In the expansion of $(a+b{)}^{n}$

${15}^{}$ term from beginning: ${T}_{15}={}^{n}{C}_{14}{a}^{n-14}{b}^{14}$
${15}^{}$ term from end: ${T}_{'}^{15}={}^{n}{C}_{14}{b}^{n-14}{a}^{14}$

$\begin{gathered} 	herefore \, \frac{{T}_{15}}{{T}_{'}^{15}}=\frac{1}{6} \ {(\frac{a}{b})}^{n-28}=\frac{1}{6} \ {({6}^{\frac{1}{3}})}^{n-28}={6}^{-1} \ \Rightarrow \, \frac{n-28}{3}=-1 \ n=25 \ 	herefore \, {}^{25}{C}_{3}=2300 \end{gathered}$

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