The given expression is:

$\sum_{n=1}^{\infty} \frac{(-1)^n 10^n (x - 10)^n}{n!}.$

This series looks like a modification of the Taylor or Maclaurin series for the exponential function. Let me analyze and simplify it.

Step-by-Step Solution:

1. General Form of the Exponential Series: Recall that the Taylor expansion for $e^x$ around $x = 0$ is: $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}.$

2. Recognizing the Structure: The given series has the term $(-1)^n 10^n (x - 10)^n$. If we rewrite it, we get: $\sum_{n=1}^\infty \frac{\left[(-1)(10)(x-10)\right]^n}{n!}.$ Let $z = -10(x - 10)$. Substituting this into the series, we have: $\sum_{n=1}^\infty \frac{z^n}{n!}.$

3. Simplifying the Summation: This summation can be related to the Taylor series of $e^z$. Recall: $e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.$ Since our series starts at $n = 1$ (not $n = 0$), we can write: $\sum_{n=1}^\infty \frac{z^n}{n!} = e^z - 1.$

4. Substitute Back $z$: Substitute $z = -10(x - 10)$ into $e^z - 1$: $\sum_{n=1}^\infty \frac{(-1)^n 10^n (x - 10)^n}{n!} = e^{-10(x - 10)} - 1.$

Final Answer:

$\sum_{n=1}^\infty \frac{(-1)^n 10^n (x - 10)^n}{n!} = e^{-10(x - 10)} - 1.$

Would you like a detailed explanation of how the substitution and simplification work? Let me know!

Related Questions:

1. What is the Taylor series of $e^x$ centered at a different point $a$?
2. How does the structure of alternating series affect convergence?
3. What are some real-world applications of exponential functions and series expansions?
4. How does the factorial in the denominator ensure the series converges for all $x$?
5. Can this series representation help in numerical approximations of $e^{-10(x - 10)} - 1$?

Tip:

When working with Taylor series, always try to identify the general pattern of coefficients to relate it to standard series expansions like $e^x$, $\sin(x)$, or $\cos(x)$. This saves time and simplifies the solution!