Answer: (a)(i) \(1+4nx+8n(n-1)x^2\). (a)(ii) \(n=29\). (b) \(\displaystyle \frac{45}{4}\).

Answer: (a)(i) \(1+4nx+8n(n-1)x^2\). (a)(ii) \(n=29\). (b) \(\displaystyle \frac{45}{4}\).

(a)(i) Using the binomial expansion,

\((1+4x)^n=1+n(4x)+\frac{n(n-1)}{2}(4x)^2+\cdots\).

Therefore the first three terms are

\(1+4nx+8n(n-1)x^2\).

(a)(ii) In

\((1+4x)^n(1-4x)\),

the coefficient of \(x^2\) comes from two contributions:

\(8n(n-1)x^2\cdot 1\) and \(4nx\cdot (-4x)\).

So the coefficient of \(x^2\) is

\(8n(n-1)-16n\).

It is given that this is 6032, so

\(8n(n-1)-16n=6032\).

Dividing by 8 gives

\(n(n-1)-2n=754\),

so

\(n^2-3n-754=0\).

Factorising,

\((n-29)(n+26)=0\).

Since \(n\) is positive,

\(n=29\).

(b) The general term is

\(\displaystyle {10\choose r}\left(\frac{x}{2}\right)^{10-r}\left(-\frac{8}{x^4}\right)^r\).

The power of \(x\) in this term is

\((10-r)-4r=10-5r\).

For the term independent of \(x\), set

\(10-5r=0\),

so \(r=2\).

The required term is therefore

\(\displaystyle {10\choose 2}\left(\frac{x}{2}\right)^8\left(-\frac{8}{x^4}\right)^2\).

This simplifies to