Question

Find the sum of the series.\\ $\sum_{n=0}^{\infty} \frac{(-1)^n 3^n x^{3n}}{n!}$\\Step 1\\We know that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.\\The series $\sum_{n=0}^{\infty} \frac{(-1)^n 3^n x^{3n}}{n!}$ can be re-written as $\sum_{n=0}^{\infty} \frac{(-3x^3)^n}{n!}$.\\Step 2\\Therefore, $\sum_{n=0}^{\infty} \frac{(-1)^n 3^n x^{3n}}{n!} = $

          Find the sum of the series.\\
$\sum_{n=0}^{\infty} \frac{(-1)^n 3^n x^{3n}}{n!}$\\Step 1\\We know that $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.\\The series $\sum_{n=0}^{\infty} \frac{(-1)^n 3^n x^{3n}}{n!}$ can be re-written as $\sum_{n=0}^{\infty} \frac{(-3x^3)^n}{n!}$.\\Step 2\\Therefore, $\sum_{n=0}^{\infty} \frac{(-1)^n 3^n x^{3n}}{n!} = $

Find the sum of the series.

∑n=0^∞((-1)^n 3^n x^3n)/(n!)
Step 1
We know that e^x = ∑n=0^∞(x^n)/(n!).
The series ∑n=0^∞((-1)^n 3^n x^3n)/(n!) can be re-written as ∑n=0^∞((-3x^3)^n)/(n!).
Step 2
Therefore, ∑n=0^∞((-1)^n 3^n x^3n)/(n!) =

Added by Philip D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Zhumagali Shomanov

04/17/2023

Step 1

Step 1: We know that the series -1n33n can be re-written as n = 0 n! * (-3)^n. Show more…

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Find the sum of the series -1n33n 0 = u n! Step 1 We know that e = iu0=u The series -133n can be re-written as n = 0 n! -3.x3 n! Step 2 Therefore, n = 0 1n33 n! X Submit Skip(you cannot come back)