# Challenging Binomial Question; O Level A Maths Group Tuition

In the expansion of $\displaystyle (x^2-\frac{1}{2x^4})^n$, in descending powers of $x$, the seventh term is independent of $x$. Find the value of $n$ and the value of this term.

Solution:

$\displaystyle\begin{array}{rcl} T_{r+1}&=&{n \choose r}(x^2)^{n-r}(-\frac{1}{2}x^{-4})^r\\ &=& {n\choose r}x^{2n-2r}(-\frac{1}{2})^r (x^{-4r})\\ &=& {n\choose r}(-\frac{1}{2})^r x^{2n-6r} \end{array}$

$r=6$ since it is the seventh term (recall $T_{r+1}$)

$2n-6r=0$ (independent of $x$ means power is 0)

$2n-36=0$

$n=18$

${18\choose 6}\times (-\frac{1}{2})^6 =290 \frac{1}{16}$ (Ans)

## Author: mathtuition88

https://mathtuition88.com/

This site uses Akismet to reduce spam. Learn how your comment data is processed.