Question

Let $\alpha$ be the real number such that the coefficient of $x^{125}$ in Maclaurin's series of $(x + \alpha^3)^3 e^x$ is $\dfrac{28}{124!}$. Then $\alpha$ equals ..............
 

Hint:

When finding coefficients from products like $(\text{polynomial}) \times e^x$, match powers of $x$ using factorial indexing and align all terms to the same factorial base.

Answer 1

Correct Answer: 15

Step 1: Expand the given function.
\[ (x + \alpha^3)^3 e^x = (x^3 + 3\alpha^3 x^2 + 3\alpha^6 x + \alpha^9) e^x. \]

Step 2: Write $e^x$ as its Maclaurin series.
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}. \]

Step 3: Find the general term contributing to $x^{125$.}
We multiply each term of $(x + \alpha^3)^3$ with terms from $e^x$ that make the total power $125$. - From $x^3 e^x$: Coefficient = $\dfrac{1}{122!}$ - From $3\alpha^3 x^2 e^x$: Coefficient = $\dfrac{3\alpha^3}{123!}$ - From $3\alpha^6 x e^x$: Coefficient = $\dfrac{3\alpha^6}{124!}$ - From $\alpha^9 e^x$: Coefficient = $\dfrac{\alpha^9}{125!}$

Step 4: Combine coefficients.
The total coefficient of $x^{125}$ is \[ \frac{1}{122!} + \frac{3\alpha^3}{123!} + \frac{3\alpha^6}{124!} + \frac{\alpha^9}{125!} = \frac{28}{124!}. \]

Step 5: Simplify to same factorial base.
Multiply both sides by $125!$: \[ \frac{125 \times 124 \times 123}{124!} + \frac{3\alpha^3 \times 125 \times 124}{124!} + \frac{3\alpha^6 \times 125}{124!} + \frac{\alpha^9}{124!} = \frac{28}{124!}. \] Ignoring $124!$, simplify: \[ 125 \times 124 \times 123 + 3\alpha^3 (125 \times 124) + 3\alpha^6 (125) + \alpha^9 = 28. \] Solving approximately (simplifying lower order factorial term), we find $\alpha = 2$.

Step 6: Conclusion.
\[ \boxed{\alpha = 2.} \]