Answer: \(n=10\).

Expand the terms needed to form \(x^2\):

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

Multiplying by \(1+x\), the coefficient of \(x^2\) is formed from two contributions:

\(1\cdot \binom{n}{2}\frac{x^2}{4}\)

and

\(x\cdot n\left(-\frac{x}{2}\right).\)

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

\(\frac14\binom{n}{2}-\frac{n}{2}.\)

This is given as \(\frac{25}{4}\), so

\(\frac14\cdot\frac{n(n-1)}{2}-\frac{n}{2}=\frac{25}{4}.\)

Multiplying by \(8\),

\(n(n-1)-4n=50.\)

Hence

\(n^2-5n-50=0.\)

Factorising,

\((n-10)(n+5)=0.\)

Since \(n\) is a positive integer,

\(n=10.\)