A ratio of the 5th term from the beginning to the 5th term from the end in the binomial ex…

Question

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $\left(2^{1/3}+\dfrac{1}{2(3)^{1/3}}\right)^{10}$ is:

Accepted Answer

$T_5=\binom{10}{4}\left(2^{1/3}\right)^6\left(\dfrac{1}{2\cdot3^{1/3}}\right)^4$; $T_{11-5+1}=T_7=\binom{10}{6}\left(2^{1/3}\right)^4\left(\dfrac{1}{2\cdot3^{1/3}}\right)^6$

$$\frac{T_5}{T_7}=\frac{\left(2^{1/3}\right)^2}{\left(\frac{1}{2\cdot3^{1/3}}\right)^2}=2^{2/3}\times4\times3^{2/3}=4(6)^{2/3}=4(36)^{1/3}:1$$

Solutions · 1