Question

$$ \begin{array}{l} P = 1 - \frac{3}{1!} + \frac{9}{2!} - \frac{27}{3!} + \ldots \ Q = 1 + \frac{4}{1!} + \frac{16}{2!} + \frac{64}{3!} + \ldots . \ R = \log _{e} 3 + \frac{(\log _{e} 3)^{2}}{2!} + \frac{(\log _{e} 3)^{3}}{3!} + \ldots . \end{array} $$

The ascending order of $P$, $Q$, $R$: A $Q$, $P$, $R$ B $P$, $R$, $Q$ C $P$, $Q$, $R$ D $R$, $P$, $Q$

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Answer

To determine the ascending order of the given series ( P ), ( Q ), and ( R ), we'll analyze each series in relation to the exponential series expansion of ( e^x ).

Analyzing Series ( P )

The series ( P ) is: $$ P = 1 - \frac{3}{1!} + \frac{9}{2!} - \frac{27}{3!} + \ldots $$

This can be identified as the expansion for ( e^{-3} ): $$ P = e^{-3} $$

Analyzing Series ( Q )

The series ( Q ) is: $$ Q = 1 + \frac{4}{1!} + \frac{16}{2!} + \frac{64}{3!} + \ldots $$

This series matches the expansion for ( e^{4} ): $$ Q = e^{4} $$

Analyzing Series ( R )

The series ( R ) is: $$ R = \log_e 3 + \frac{(\log_e 3)^2}{2!} + \frac{(\log_e 3)^3}{3!} + \ldots $$

This fits the exponential expansion but without the initial 1, so: $$ R = e^{\log_e 3} - 1 = 3 - 1 = 2 $$ (Note that ( e^{\log_e 3} = 3 ))

Determining Ascending Order

To compare these values, note:

  • ( e^{-3} \approx 0.050 )
  • ( e^4 \approx 54.598 )
  • ( R = 2 )

Thus, the ascending order is: $$ P \approx 0.050 < R = 2 < Q \approx 54.598 $$

Conclusion

The ascending order of ( P ), ( Q ), and ( R ) is: ( P, R, Q )

Final Answer: B


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