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)

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