JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

The value of

1 + \frac{{{{\left( {{{\log }_e}n} \right)}^2}}}{{2\;!}} + \frac{{{{\left( {{{\log }_e}n} \right)}^4}}}{{4\;!}} + \ldots \mathrm{is}

View answer and explanation

Correct option: C

\frac{{n + {n^{ - 1}}}}{2}

Option C (

\frac{{n + {n^{ - 1}}}}{2}

) is correct because the given series is the expansion of

\frac{{{e^{{{\log }_e}n}} + {e^{ - {{\log }_e}n}}}}{2}

, which simplifies to

\frac{{n + {n^{ - 1}}}}{2}

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