Question

Find the $6^{th}$ term of ${\left(1+\frac{x}{2}\right)}^{-5}$

Answer 1

Solution -

$6^{th}$ term of $(1+\frac{x}{2}{)}^{-5}$ expansion

$6^{th}$ term $=\frac{-n(n+1)(n+2)(n+3)(n+4)}{1\times 2\times 3\times 4\times 5}(\frac{x}{2}{)}^5$

$=\frac{-n(n+1)(n+2)(n+3)(n+4)x^5}{3840}$