JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Understanding +4 | -1

1 + \frac{{{{\left( {{{\log }_e}n} \right)}^2}}}{{2!}} + \frac{{{{\left( {{{\log }_e}n} \right)}^4}}}{{4!}} + ...

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}\left( {n + {n^{ - 1}}} \right)

Option C (

\frac{1}{2}\left( {n + {n^{ - 1}}} \right)

) is correct because the given series is the expansion of

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

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