Consecutive Terms Sum
Consider the sequence defined by $ a_k=\frac 1{k^2+k} $ for $ k\ge 1 $. Given that $ a_m+a_{m+1}+\cdots+a_{n-1}=\frac{1}{29} $, for positive integers $ m $ and $ n $ with $ m<n $, find $ m+n $.
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Solution
Note that \[a_k = \frac{1}{k^2+k} = \frac{1}{k} - \frac{1}{k+1}\]for each $ k $. Thus, the sum telescopes: \[\begin{aligned} a_m + a_{m+1} + \dots + a_{n-1} & = \left(\frac{1}{m} - \frac{1}{m+1}\right) + \left(\frac{1}{m+1} - \frac{1}{m+2}\right) + \dots + \left(\frac{1}{n-1} - \frac{1}{n}\right) \\ &= \frac{1}{m} - \frac{1}{n}.\end{aligned}\]Therefore, we have the equation $ 1/m - 1/n = 1/29 $. Multiplying by $ 29mn $ on both sides, we have $ 29n - 29m = mn, $ or $ mn + 29m - 29n = 0 $. Subtracting $ 29^2 $ from both sides, we get \[(m-29)(n+29) = -29^2.\]Since $ 29 $ is prime and $ 0 < m < n, $ the only possibility is $ m-29 = -1 $ and $ n+29 = 841, $ which gives $ m = 28 $ and $ n = 812 $. Thus, $ m+n=28+812=\boxed{840} $.
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