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CSS Pure Mathematics QUESTION #4101
Question 1
The Maclaurin series for \(f(Z)=Z^2e^{3Z}\), using \(e^{3Z}=\displaystyle\sum_{n=0}^{\infty}\dfrac{3^nZ^n}{n!}\), begins as:
  • \(Z^2+3Z^3+\dfrac{9}{2}Z^4+\cdots\)✔️
  • \(Z^2+3Z^3+\dfrac{9}{2}Z^4+\dfrac{9}{2}Z^5+\cdots\)
  • \(1+3Z+\dfrac{9Z^2}{2}+\cdots\)
  • \(Z^2+Z^3+Z^4+\cdots\)
Correct Answer Explanation
\(Z^2e^{3Z}=Z^2\sum_{n=0}^\infty\dfrac{3^nZ^n}{n!}=Z^2+3Z^3+\dfrac{3^2}{2!}Z^4+\dfrac{3^3}{3!}Z^5+\cdots = Z^2+3Z^3+\dfrac{9}{2}Z^4+\dfrac{9}{2}Z^5+\cdots\).