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UPSC NDA I 2018 Mathematics Solved Paper

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Question : 33 of 120

Marks: +1, -0

If n = (2017)! then what is

\frac{1}{log_{2}n}+\frac{1}{log_{3}n}+\frac{1}{log_{4}n}+…+\frac{1}{log_{2017}n}

equal to?

0
1
n/2
n

Solution:

Concept:

Use the logarithm property

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

and the product rule for logs to simplify the sum.

Explanation:

Given

n = (2017)!

.

We need

\frac{1}{\log_{2}n} + \frac{1}{\log_{3}n} + \cdots + \frac{1}{\log_{2017}n}

.

Apply

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

: each term

\frac{1}{\log_k n} = \log_n k

for

k = 2, 3, \dots, 2017

.

So the sum becomes

\log_n 2 + \log_n 3 + \cdots + \log_n 2017

.

Using

\log_n a + \log_n b = \log_n (ab)

, the sum equals

\log_n (2 \times 3 \times \cdots \times 2017) = \log_n (2017!)

.

Since

n = (2017)!

, we have

\log_n n = 1

.

Thus the entire sum equals

1

.

Answer:

1

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