The coefficient of {x^r} in the expansion of | vedclass

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(C) The expansion of ${e^{e^x}}$ is given by:${e^{e^x}} = e \cdot e^{e^x - 1} = e \cdot \sum_{n=0}^{\infty} \frac{(e^x - 1)^n}{n!}$.Alternatively,usin

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$\frac{1}{r!} \left[ \frac{1^r}{1!} + \frac{2^r}{2!} + \frac{3^r}{3!} + \dots \right]$

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(C) The expansion of ${e^{e^x}}$ is given by:${e^{e^x}} = e \cdot e^{e^x - 1} = e \cdot \sum_{n=0}^{\infty} \frac{(e^x - 1)^n}{n!}$.Alternatively,using the series expansion of ${e^y}$ where $y = e^x$:${e^{e^x}} = \sum_{n=0}^{\infty} \frac{(e^x)^n}{n!} = \sum_{n=0}^{\infty} \frac{e^{nx}}{n!}$.Expanding each term $e^{nx}$ as a power series in $x$:${e^{nx}} = \sum_{r=0}^{\infty} \frac{(nx)^r}{r!} = \sum_{r=0}^{\infty} \frac{n^r x^r}{r!}$.Substituting this back into the summation:${e^{e^x}} = \sum_{n=0}^{\infty} \frac{1}{n!} \left( \sum_{r=0}^{\infty} \frac{n^r x^r}{r!} \right)$.To find the coefficient of ${x^r}$,we collect the terms containing ${x^r}$:Coefficient of ${x^r} = \sum_{n=1}^{\infty} \frac{1}{n!} \cdot \frac{n^r}{r!} = \frac{1}{r!} \sum_{n=1}^{\infty} \frac{n^r}{n!}$.Thus,the coefficient is $\frac{1}{r!} \left[ \frac{1^r}{1!} + \frac{2^r}{2!} + \frac{3^r}{3!} + \dots \right]$.

The coefficient of ${x^r}$ in the expansion of ${e^{e^x}}$ is

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