In the expansion of frac{1 + x}{1!} + frac{(1 | vedclass

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(C) The given series is $S = \frac{1 + x}{1!} + \frac{(1 + x)^2}{2!} + \frac{(1 + x)^3}{3!} + \dots \infty$.We know that the exponential series is $e^

Vedclass · Accepted Answer

$\frac{e}{n!}$

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(C) The given series is $S = \frac{1 + x}{1!} + \frac{(1 + x)^2}{2!} + \frac{(1 + x)^3}{3!} + \dots \infty$.We know that the exponential series is $e^y = 1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots$.Comparing this,we have $S = e^{1 + x} - 1$.$S = e \cdot e^x - 1$.Substituting the expansion of $e^x = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dots$,we get:$S = e \left( 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dots \right) - 1$.$S = (e - 1) + ex + e\frac{x^2}{2!} + \dots + e\frac{x^n}{n!} + \dots$.Therefore,the coefficient of $x^n$ is $\frac{e}{n!}$.

In the expansion of $\frac{1 + x}{1!} + \frac{(1 + x)^2}{2!} + \frac{(1 + x)^3}{3!} + \dots$,the coefficient of $x^n$ will be

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