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: Find Maclaurin series

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

$$(x + \alpha^3)e^x = x \cdot e^x + \alpha^3 \cdot e^x$$

$$= x \sum_{n=0}^{\infty} \frac{x^n}{n!} + \alpha^3 \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

$$= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} + \sum_{n=0}^{\infty} \frac{\alpha^3 x^n}{n!}$$

Step 2: Find coefficient of $x^{125}$

For the term $x^{125}$:

From $\sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}$:

• We need $n + 1 = 125$, so $n = 124$
• Coefficient: $\frac{1}{124!}$

From $\sum_{n=0}^{\infty} \frac{\alpha^3 x^n}{n!}$:

• We need $n = 125$
• Coefficient: $\frac{\alpha^3}{125!}$

Total coefficient of $x^{125}$: $$\frac{1}{124!} + \frac{\alpha^3}{125!} = \frac{1}{124!} + \frac{\alpha^3}{125 \cdot 124!} = \frac{125 + \alpha^3}{125 \cdot 124!}$$

Step 3: Set equal to given coefficient

$$\frac{125 + \alpha^3}{125 \cdot 124!} = \frac{28}{124!}$$

Multiply both sides by $124!$: $$\frac{125 + \alpha^3}{125} = 28$$

$$125 + \alpha^3 = 28 \times 125 = 3500$$

$$\alpha^3 = 3500 - 125 = 3375$$

$$\alpha = \sqrt[3]{3375} = \sqrt[3]{27 \times 125} = \sqrt[3]{27} \times \sqrt[3]{125} = 3 \times 5 = 15$$

Answer: $\alpha = 15$