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Logarithmic Function

The coefficie...

Question

The coefficient of $x^{n}$ in the series $1 + \frac{(2 + b x)}{1!} + \frac{(2 + b x)^{2}}{2!} + . . . + \frac{(2 + b x)^{n}}{n!}$ is $\frac{e^{a} b^{n}}{n!}$ . Find $a$

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Solution

Let $a + b x = y$ then
$1 + \frac{a + b x}{1!} + \frac{(a + b x)^{2}}{2!} + \frac{(a + b x)^{3}}{3!} + . . . \infty$
$= 1 + \frac{y}{1!} + \frac{y^{2}}{2!} + \frac{y^{3}}{3!} + . . . \infty$
$= e^{y}$
$= e^{a + b x}$
$= e^{a} . e^{b x}$
$= e^{a} (1 + b x + \frac{(b x)^{2}}{2!} + . . . + \frac{(b x)^{n}}{n!} + . . .)$
$= e^{a} (1 + b x + \frac{b^{2} x^{2}}{2!} + . . . + \frac{b^{3} x^{n}}{n!} + . . .)$
Hence, coefficient of $x^{n} =$ $\frac{e^{a} b^{n}}{n!}$
$\Rightarrow a = 2$

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