Question

Let the Taylor polynomial of degree 20 for \( \frac{1}{(1-x)^3} \) at \( x = 0 \) be

\[ \sum_{n=0}^{20} a_n x^n \]

Then \( a_{15} \) equals

• A. 136
• B. 120
• C. 60
• D. 272

Hint:

To find coefficients in a Taylor series, use the general formula for the Taylor series and apply it to the specific function you are working with.

Answer 1

The Correct Option is A

Step 1: The Taylor series for \( \frac{1}{(1-x)^3} \). 
The function \( \frac{1}{(1-x)^3} \) can be expanded into a Taylor series at \( x = 0 \). The general form of the Taylor series for this function is:

\[ \frac{1}{(1-x)^3} = \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \]

We need to find the coefficient \( a_{15} \), which is the 15th term of the series. 
Step 2: Finding \( a_{15} \). 
For \( n = 15 \), the coefficient is:

\[ a_{15} = \binom{15+2}{2} = \binom{17}{2} = 136 \]

Step 3: Conclusion. 
The correct answer is (B) 136.