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UPSC NDA I 2021 Mathematics Solved Paper

Question : 54 of 120

Marks: +1, -0

n = 100

\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \cdots + \frac{1}{\log_{100} n}

Solution:

Concept:

Use the change of base rule:

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

, and then combine the sum of logs into a single log of a product.

Explanation:

Start with the given sum:

\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \cdots + \frac{1}{\log_{100} n}

Apply the change of base rule: each term

\frac{1}{\log_k n}

becomes

\log_n k

So the sum becomes

\log_n 2 + \log_n 3 + \log_n 4 + \cdots + \log_n 100

Since

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

, we combine all terms:

\log_n (2 \times 3 \times 4 \times \cdots \times 100)

The product

2 \times 3 \times 4 \times \cdots \times 100

is exactly

100!

. Also,

n = 100!

is given.

Thus the expression simplifies to

\log_n n = \log_{100!} 100!

, which equals

1

Answer:

1

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