e^{left( {x - frac{1}{2}{(x - 1)}^2 + frac{1} | vedclass

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(D) The given expression is $E = e^{\left( {x - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \frac{1}{4}{(x - 1)}^4 + \dots} \right)}$.We can rew

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$xe$

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(D) The given expression is $E = e^{\left( {x - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \frac{1}{4}{(x - 1)}^4 + \dots} \right)}$.We can rewrite the exponent by adding and subtracting $1$:$E = e^{\left( {(x - 1) - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \dots} \right) + 1}$.Using the logarithmic series expansion $\log(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,where $t = x - 1$,we get:$E = e^{\log(1 + (x - 1)) + 1} = e^{\log x + 1}$.Using the property $e^{a+b} = e^a \cdot e^b$,we have:$E = e^{\log x} \cdot e^1 = x \cdot e = xe$.

$e^{\left( {x - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \frac{1}{4}{(x - 1)}^4 + \dots} \right)}$ is equal to

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