Question

Find the middle term(s) in the expansion of

${(1-\frac{x^2}{2})}^{14}$

Answer 1

${(1-\frac{x^2}{2})}^{14}$
Here, $n$ is even, therefore middle term is $\left(\frac{14+2}{2}\right)$th term.
It means $T_8$ is middle term
$T_8={}^{14}{\text{C}}_7{(-\frac{x^2}{2})}^7=-\frac{429}{16}{x}^{14}$