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CSS Pure Mathematics QUESTION #4108
Question 1
The Maclaurin series \(\displaystyle f(Z)=Z^2e^{3Z}=\sum_{n=0}^{\infty}\dfrac{3^nZ^{n+2}}{n!}\). The coefficient of \(Z^5\) in this series is:
  • \(\dfrac{27}{6}\)✔️
  • \(\dfrac{9}{2}\)
  • \(9\)
  • \(\dfrac{81}{24}\)
Correct Answer Explanation
The coefficient of \(Z^5\) corresponds to \(n+2=5\Rightarrow n=3\): coefficient \(=\dfrac{3^3}{3!}=\dfrac{27}{6}=\dfrac{9}{2}\). Verify: \(\dfrac{27}{6}=\dfrac{9}{2}\) — both options 1 and 2 are the same value. The coefficient is \(\dfrac{9}{2}\).