Question

Find the middle term of ${(1-\frac{x^2}{2})}^{14}$.

Answer 1

The number of terms in the given expression will be $14+1=15\therefore$ the middle term will be the $8^{th}$ term
By binomial expansion, the $8^{th}$ term will be ${}^{14}{C}_7{(-\frac{x^2}{2})}^7=\frac{-429}{16}{x}^{14}$