Summation

Problem: [ARML 2003]

Find the value of $x + y$ where

$$2^x 3^y = \prod_{i = 1}^n (24^{\frac{1}{i+1} + \frac{1}{i+1} + \dots + \frac{1}{n+1}})^i$$

Solution:

$$\ \ \ \ \ \ \ \ S = \sum_{i = 1}^n \sum_{j = i+1}^{n+1} \frac{i}{j}
\ = \sum_{j = 2}^{n+1} \sum_{i = 1}^{j-1} \frac{i}{j}
\ \ = \sum_{j = 2}^{n+1} \frac{1}{j} \sum_{i = 1}^{j-1} i \\
= \sum_{j = 2}^{n+1} \frac{1}{j} \cdot \frac{j(j-1)}{2}
\ = \frac{1}{2} \sum_{j = 2}^{n+1} j - 1 \\ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \sum_{j = 1}^n j
\ = \frac{1}{2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{4} $$

Hence, $x + y = 4S = n(n+1)$