In the expansion of ([3]2+1[3]3)^n, if the ratio of the 15th term from the beginning to th…

Question

In the expansion of $\left(\sqrt[3]2+\dfrac{1}{\sqrt[3]{3}}\right)^n$, if the ratio of the $15$th term from the beginning to the $15$th term from the end is $\dfrac16$, then the value of ${}^nC_3$ is:

Accepted Answer

The 15th term from the beginning: $T_{15}=\binom{n}{14}a^{n-14}b^{14}$. The 15th term from the end (of $n+1$ terms total): $T'_{15}=\binom{n}{14}b^{n-14}a^{14}$.

$$\frac{T_{15}}{T'_{15}}=\left(\frac ab\right)^{n-28}=\frac16$$

With $a=2^{1/3},\ b=3^{-1/3}$, $\dfrac ab=6^{1/3}$, so $\left(6^{1/3}\right)^{n-28}=6^{-1}\Rightarrow\dfrac{n-28}{3}=-1\Rightarrow n=25$.

$$\binom{25}{3}=2300$$

Solutions · 1