The coefficient of $x^{n}$ in the expansion of $e^{2 x + 3}$ is

\frac{2^{n}}{n!}

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\frac{e^{3} . 2^{n}}{n!}

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\frac{e^{2} . 3^{n}}{n!}

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none of these

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Solution

The correct option is B $\frac{e^{3} . 2^{n}}{n!}$
Given, $e^{2 x + 3}$ $= e^{2 x} e^{3}$
$= e^{3} (1 + 2 x + \frac{(2 x)^{2}}{2!} + . . . \infty)$
Hence the general term will be
$\frac{(2 x)^{n} e^{3}}{n!}$

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Similar questions

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(b)

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(c)

\frac{(2 n)!}{(2 n - 1)! (2 n + 1)!}

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The coefficient of two consecutive terms in the

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The coefficient of xn in the expansion of e2x+3 is

The correct option is B e3.2nn!Given, e2x+3 =e2xe3=e3(1+2x+(2x)22!+...∞)Hence the general term will be(2x)ne3n!

The coefficient of $x^{n}$ in the expansion of $e^{2 x + 3}$ is

The correct option is B $\frac{e^{3} . 2^{n}}{n!}$
Given, $e^{2 x + 3}$ $= e^{2 x} e^{3}$
$= e^{3} (1 + 2 x + \frac{(2 x)^{2}}{2!} + . . . \infty)$
Hence the general term will be
$\frac{(2 x)^{n} e^{3}}{n!}$