Question

If $f(x) = \left( \frac{x}{2} \right)^{10}$, then $f(1) + \frac{f'(1)}{1!} + \frac{f''(1)}{2!} + \frac{f'''(1)}{3!} + \dots + \frac{f^{(10)}(1)}{10!}$ is equal to:

• A. $1$
• B. $10$
• C. $11$
• D. $512$
• E. $1024$

Hint:

In any polynomial $P(x)$ of degree $n$, the sum $\sum_{k=0}^{n} \frac{P^{(k)}(a)}{k!}$ is always equal to $P(a+1)$. This is a direct consequence of the Taylor expansion where the step size $h = 1$.

Answer 1

The Correct Option is A

Concept: The expression given is the Taylor series expansion of $f(x)$ centered at $a = 1$. The general Taylor series for a function $f(x)$ about $x = a$ is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n \]

Step 1: Identify the structure of the series.
The provided sum is: \[ S = \frac{f^{(0)}(1)}{0!} + \frac{f^{(1)}(1)}{1!} + \frac{f^{(2)}(1)}{2!} + \dots + \frac{f^{(10)}(1)}{10!} \] Notice that this matches the Taylor series for $f(x)$ at $a = 1$ if we set $(x - 1) = 1$.

Step 2: Determine the value of $x$.
To make $(x - 1) = 1$, we must have $x = 2$. Therefore, the sum $S$ is equivalent to $f(2)$.

Step 3: Calculate $f(2)$.
Given $f(x) = \left( \frac{x}{2} \right)^{10}$: \[ f(2) = \left( \frac{2}{2} \right)^{10} = 1^{10} = 1 \]