The value of {1}{{{{ }₃e}} + {1}{{{{ }₃{e²}} + {1}{{{{ }₃{e^4}}} + up to infinite terms is

Correct Answer is : loge9log_{e} 9loge9

Correct Answer is : loge9log_{e} 9loge9

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 25 of 120

Marks: +1, -0

The value of

\frac{1}{{{{\log }_3}e}} + \frac{1}{{{{\log }_3}{e^2}}} + \frac{1}{{{{\log }_3}{e^4}}} + \ldots

up to infinite terms is

Solution:

Concept:

Transform each term using logarithm properties to form an infinite geometric series.

Explanation:

Rewrite each denominator:

{\log_3}(e^2) = 2{\log_3}e

{\log_3}(e^4) = 4{\log_3}e

, etc.

Thus

\frac{1}{{\log_3}e^2} = \frac{1}{2{\log_3}e}

\frac{1}{{\log_3}e^4} = \frac{1}{4{\log_3}e}

, and so on.

The series becomes

\frac{1}{{\log_3}e} \left[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \right]

The bracket is a geometric series with first term

1

and common ratio

\frac{1}{2}

Its sum is

\frac{1}{1 - \frac12} = 2

Thus the whole expression equals

\frac{2}{{\log_3}e}

Using

\frac{1}{{\log_a}b} = {\log_b}a

, we get

2\,{\log_e}3 = {\log_e}(3^2) = {\log_e}9

Answer:

{\log_e}9

(option A).

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