0606 P13 - Jun 2023 - Q5 - 6 marks
7677
(a) Find the first three terms in the expansion of
\(\left(x^2-\frac4{x^2}\right)^{10}\)
in descending powers of \(x\). Give each term in its simplest form.
(b) Hence find the coefficient of \(x^{16}\) in the expansion of
\(\left(x^2-\frac4{x^2}\right)^{10}\left(x^2+\frac2{x^2}\right)^2.\)
Solution
Answer: (a) \(x^{20}-40x^{16}+720x^{12}\); (b) \(564\).
Use the binomial theorem carefully, keeping track of the power of each factor in the general term.
(a) Use the binomial expansion of
\(\left(x^2-\frac4{x^2}\right)^{10}.\)
The first term is
\((x^2)^{10}=x^{20}.\)
The second term is
\(\binom{10}{1}(x^2)^9\left(-\frac4{x^2}\right) =10x^{18}\left(-\frac4{x^2}\right) =-40x^{16}.\)
The third term is
\(\binom{10}{2}(x^2)^8\left(-\frac4{x^2}\right)^2 =45x^{16}\frac{16}{x^4} =720x^{12}.\)
So the first three terms are
\(x^{20}-40x^{16}+720x^{12}.\)
(b) Now
\(\left(x^2+\frac2{x^2}\right)^2 =x^4+4+\frac4{x^4}.\)
To form \(x^{16}\), use:
\(x^{20}\cdot\frac4{x^4},\qquad -40x^{16}\cdot4,\qquad 720x^{12}\cdot x^4.\)
The coefficient of \(x^{16}\) is therefore
\(4-160+720=564.\)
