sum_(n=1)^(oo)(3^(n)n^(2))/(n!)

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Recognize Power Series Expansion: Recognize the series as a power series expansion. The given series is ∑_(n=1)^(∞)(3^n * n^2)/(n!), which resembles the expansion of e^x, where e^x = ∑_(n=0)^(∞)(x^n)/(n!). However, our series has additional factors of 3^n and n^2.Compare with e^(3x) Series: Compare the given series with the expansion of e^(3x). The series for e^(3x) is ∑_(n=0)^(∞)(3x)^n/(n!) = ∑_(n=0)^(∞)(3^n * x^n)/(n!). We notice that if we differentiate e^(3x) with respect to x, we introduce a factor of n in the numerator, which gets us closer to the given series.First Derivative of e^(3x): Differentiate e^(3x) with respect to x once. The first derivative of e^(3x) with respect to x is d/dx[e^(3x)] = 3e^(3x). The series representation after differentiating once is ∑_(n=0)^(∞)(3^n * x^(n-1) * n)/(n!).Second Derivative of e^(3x): Differentiate e^(3x) with respect to x a second time. The second derivative of e^(3x) with respect to x is d^2/dx^2[e^(3x)] = 9e^(3x). The series representation after differentiating twice is ∑_(n=0)^(∞)(3^n * x^(n-2) * n * (n-1))/(n!).Evaluate Second Derivative at x=1: Evaluate the second derivative of e^(3x) at x = 1. By setting x = 1 in the series representation of the second derivative, we get ∑_(n=0)^(∞)(3^n * 1^(n-2) * n * (n-1))/(n!) = ∑_(n=0)^(∞)(3^n * n * (n-1))/(n!). This series is almost the same as the given series, except it starts at n = 0 and includes an (n-1) instead of n^2.Adjust Series to Match Given Series: Adjust the series to match the given series. We need to start the series from n = 1 and replace (n-1) with n^2. To do this, we can add and subtract the term for n = 0 from the series we obtained in Step 5. The term for n = 0 is 0 (since 0 * (0-1) = 0), so it does not contribute to the sum. We can then start the series from n = 1 as required.Write Final Series and Evaluate Sum: Write the final series and evaluate the sum. The final series, starting from n = 1, is ∑_(n=1)^(∞)(3^n * n^2)/(n!) = 9e^(3) - 9, because the second derivative of e^(3x) evaluated at x = 1 is 9e^(3), and we subtract 9 to account for the n = 1 term in the series of the second derivative, which is not present in the original series.

$\sum_{n=1}^{\infty}\frac{3^{n}n^{2}}{n!}$

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$\sum_{n=1}^{\infty}\frac{3^{n}n^{2}}{n!}$