Challenging Binomial Question; O Level A Maths Group Tuition

Question: (Broadrick Sec Prelim Add Math Paper 1 2010, Q8b)

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

http://mathtuition88.com View all posts by mathtuition88

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