Question

Suppose $f(n)=log_2(3).log_3(4).log_4(5)....log_{n-1}(n)$ then the sum $\sum_{k=2}^{100} f(2^k)$ equals

• A. 5010
• B. 5050
• C. 5100
• D. 5049

Answer 1

Answer: (D)

5049

Answer 2

The function $f(n)$ can be simplified using the change of base formula for logarithms: $f(n) = \log_2(3) \cdot \log_3(4) \cdot \log_4(5) \cdots \log_{n-1}(n)$ Using $\log_a(b) = \frac{\ln(b)}{\ln(a)}$, we get: $f(n) = \frac{\ln(3)}{\ln(2)} \cdot \frac{\ln(4)}{\ln(3)} \cdot \frac{\ln(5)}{\ln(4)} \cdots \frac{\ln(n)}{\ln(n-1)}$ This is a telescoping product, which simplifies to: $f(n) = \frac{\ln(n)}{\ln(2)} = \log_2(n)$

Now we need to compute the sum $\sum_{k=2}^{100} f(2^k)$: $\sum_{k=2}^{100} f(2^k) = \sum_{k=2}^{100} \log_2(2^k)$ Using the property $\log_a(a^x) = x$: $\sum_{k=2}^{100} \log_2(2^k) = \sum_{k=2}^{100} k$

This is the sum of an arithmetic series $2 + 3 + \cdots + 100$. The sum can be calculated as: $\sum_{k=2}^{100} k = \left(\sum_{k=1}^{100} k\right) - 1$ The sum of the first 100 integers is $\frac{100(100+1)}{2} = \frac{100 \times 101}{2} = 5050$. Therefore, the required sum is $5050 - 1 = 5049$.