Rodriguez Lang
06/08/2024 · Primary School
\( \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 4 } { n ! } \)
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Tutor-Verified Answer
Step-by-step Solution
To evaluate the series
\[
\sum_{n=0}^{\infty} \frac{(-1)^n 4}{n!},
\]
we can factor out the constant \(4\) from the summation:
\[
4 \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}.
\]
The series
\[
\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}
\]
is the Taylor series expansion for \(e^{-x}\) evaluated at \(x = 1\):
\[
e^{-1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}.
\]
Thus, we have:
\[
\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = e^{-1}.
\]
Substituting this back into our expression, we get:
\[
4 \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} = 4 e^{-1}.
\]
Therefore, the value of the original series is:
\[
\boxed{4 e^{-1}}.
\]
Quick Answer
The value of the series is \( 4 e^{-1} \).
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