To find the coefficient of \(x^7\) in the expansion of \(\left( px^2 + \frac{4}{p}x \right)^5\), we use the binomial theorem. The general term in the expansion is given by:
\(\binom{5}{r} (px^2)^{5-r} \left( \frac{4}{p}x \right)^r\)
We need the power of \(x\) to be 7, so:
\(2(5-r) + r = 7\)
\(10 - 2r + r = 7\)
\(10 - r = 7\)
\(r = 3\)
Substitute \(r = 3\) into the general term:
\(\binom{5}{3} (px^2)^{5-3} \left( \frac{4}{p}x \right)^3\)
\(= \binom{5}{3} (px^2)^2 \left( \frac{4}{p}x \right)^3\)
\(= \binom{5}{3} p^2 x^4 \cdot \frac{64}{p^3} x^3\)
\(= \binom{5}{3} \cdot \frac{64p^2}{p^3} x^7\)
\(= 10 \cdot \frac{64}{p} x^7\)
The coefficient of \(x^7\) is \(\frac{640}{p}\), and we know this equals 1280:
\(\frac{640}{p} = 1280\)
\(640 = 1280p\)
\(p = \frac{640}{1280} = \frac{1}{2}\)
