1 + frac{a - b}{a} + frac{1}{2!} left( frac{a | vedclass

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Question

(C) The given series is of the form $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = e^x$,where $x = \frac{a - b}{a}$.Substituting $x = \frac{a - b}

Vedclass · Accepted Answer

$\frac{e}{e^{b/a}}$

Vedclass · Answer

(C) The given series is of the form $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots = e^x$,where $x = \frac{a - b}{a}$.Substituting $x = \frac{a - b}{a}$ into the exponential series formula:$1 + \frac{a - b}{a} + \frac{1}{2!} \left( \frac{a - b}{a} \right)^2 + \frac{1}{3!} \left( \frac{a - b}{a} \right)^3 + \dots = e^{(a - b)/a}$.Simplifying the exponent:$e^{(a - b)/a} = e^{1 - b/a}$.Using the property of exponents $e^{m - n} = \frac{e^m}{e^n}$:$e^{1 - b/a} = e^1 \cdot e^{-b/a} = \frac{e}{e^{b/a}}$.Thus,the correct option is $C$.

$1 + \frac{a - b}{a} + \frac{1}{2!} \left( \frac{a - b}{a} \right)^2 + \frac{1}{3!} \left( \frac{a - b}{a} \right)^3 + \dots \infty = $

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