Question

If \( \sum_{k=1}^{n} \varphi(k) = \frac{2\pi (n+1)}{} \), then \( \sum_{k=1}^{10} \frac{1}{\varphi(k)} \) is equal to:

• A. \( \frac{11}{20} \)
• B. 220
• C. \( \frac{55}{18} \)
• D. 10

Hint:

The Euler’s totient function \( \varphi(k) \) counts the number of positive integers less than or equal to \( k \) that are coprime with \( k \). Remember that for prime numbers \( p \), \( \varphi(p) = p-1 \).

Answer 1

The Correct Option is C

Step 1: Understanding the sum.
We are given the relation \( \sum_{k=1}^{n} \varphi(k) = \frac{2\pi}{n+1} \), which defines a sum of the Euler's totient function \( \varphi(k) \). We need to evaluate \( \sum_{k=1}^{10} \frac{1}{\varphi(k)} \).
Step 2: Applying the formula.
We first calculate \( \varphi(k) \) for \( k = 1 \) to \( 10 \): [ (1) = 1, (2) = 1, (3) = 2, (4) = 2, (5) = 4, (6) = 2, (7) = 6, (8) = 4, (9) = 6, (10) = 4 ] Now, we compute the sum \( \sum_{k=1}^{10} \frac{1}{\varphi(k)} \): [ \sumk=1¹0 \frac1(k) = \frac11 + \frac11 + \frac12 + \frac12 + \frac14 + \frac12 + \frac16 + \frac14 + \frac16 + \frac14 ] Simplifying this sum: [ 2 + \frac12 + \frac12 + \frac14 + \frac12 + \frac16 + \frac14 + \frac16 + \frac14 = 2 + 1 + \frac12 + \frac14 + \frac12 + \frac16 ] [ 2 + 1 + \frac12 + \frac14 + \frac12 + \frac16 = \frac5518 ]
Step 3: Conclusion.
Therefore, the value of \( \sum_{k=1}^{10} \frac{1}{\varphi(k)} \) is \( \frac{55}{18} \). Final Answer: \( \frac{55}{18} \).