The coefficient of x^{6} in the expansion of | vedclass

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(C) The expansion of $e^{z}$ is given by $e^{z} = \sum_{n=0}^{\infty} \frac{z^{n}}{n!}$.Substituting $z = 2x$,we get $e^{2x} = \sum_{n=0}^{\infty} \fr

Vedclass · Accepted Answer

$\frac{4}{45}$

Vedclass · Answer

(C) The expansion of $e^{z}$ is given by $e^{z} = \sum_{n=0}^{\infty} \frac{z^{n}}{n!}$.Substituting $z = 2x$,we get $e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^{n}}{n!}$.To find the coefficient of $x^{6}$,we look at the term where $n=6$:$\frac{(2x)^{6}}{6!} = \frac{2^{6} \cdot x^{6}}{720}$.Calculating the value: $\frac{64}{720} = \frac{64 \div 16}{720 \div 16} = \frac{4}{45}$.Thus,the coefficient of $x^{6}$ is $\frac{4}{45}$.

The coefficient of $x^{6}$ in the expansion of $e^{2x}$ is:

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