Question

If the sum of the first 21 terms of the sequence \( \ln \frac{a}{b}, \ln \frac{a^2}{b^3}, \ln \frac{a^3}{b^5}, \dots \) is \( \ln \frac{a^m}{b^n} \), then the value of \( m + n \) is ________

Hint:

Always check if a sequence of logarithmic terms forms an arithmetic progression. \(T_2 - T_1 = (\ln a^2 - \ln b^3) - (\ln a - \ln b) = \ln a - 2 \ln b\). \(T_3 - T_2 = \ln a - 2 \ln b\). It is an AP. You can use the AP sum formula \(S_N = \frac{N}{2}(2T_1 + (N-1)d)\) as an alternative way to solve.

Answer 1

Correct Answer: 672

Step 1: Understanding the Concept:
This problem requires summing a sequence of logarithmic terms. This can be approached either by using logarithm properties (\(\sum \ln(x_i) = \ln(\prod x_i)\)) or by recognizing that the sequence of logs is an arithmetic progression.
Step 2: Key Formula or Approach:
The k-th term of the sequence is \( T_k = \ln\left(\frac{a^k}{b^{2k-1}}\right) \). The sum of the first 21 terms is \( S_{21} = \sum_{k=1}^{21} T_k = \ln\left( \prod_{k=1}^{21} \frac{a^k}{b^{2k-1}} \right) \). We need to find the sum of the exponents for 'a' and 'b'. Sum of powers of 'a': \( \sum_{k=1}^{21} k = \frac{21(21+1)}{2} \). Sum of powers of 'b': \( \sum_{k=1}^{21} (2k-1) \), which is the sum of the first 21 odd numbers.
Step 3: Detailed Explanation:
The sum of the sequence is given by: \[ S_{21} = \ln\left( \frac{a^1}{b^1} \cdot \frac{a^2}{b^3} \cdot \frac{a^3}{b^5} \cdots \frac{a^{21}}{b^{41}} \right) \] \[ S_{21} = \ln\left( \frac{a^{1+2+3+\dots+21}}{b^{1+3+5+\dots+41}} \right) \] Let's calculate the sum of the exponents. For 'a', the sum is an arithmetic series: \( m = \sum_{k=1}^{21} k = \frac{21(22)}{2} = 231 \).
For 'b', the sum is the sum of the first 21 odd numbers: \( n = \sum_{k=1}^{21} (2k-1) = 21^2 = 441 \).
So, the sum is \( S_{21} = \ln\left(\frac{a^{231}}{b^{441}}\right) \).
By comparing this with \( \ln\left(\frac{a^m}{b^n}\right) \), we find \( m = 231 \) and \( n = 441 \).
The value of \( m+n \) is \( 231 + 441 = 672 \).
Step 4: Final Answer:
Based on direct calculation, the value of \(m+n\) is 672.