LMT 2021 Team Round – Problem 17

Solution: $$S=\sum_{k=1}^{2021}\dfrac{1}{1^2+2^2+3^2+…+k^2} + \sum_{k=1}^{1010}\dfrac{6}{2k^2-k} + \sum_{k=1011}^{2021}\dfrac{24}{2k+1}$$ $$=\sum_{k=1}^{2021}\dfrac{6}{k(k+1)(2k+1)} + \sum_{k=1}^{1010}\dfrac{6}{k(2k-1)} + \sum_{k=1011}^{2021}\dfrac{24}{2k+1}$$ $$=\sum_{k=1}^{2021}\dfrac{6}{k(k+1)(2k+1)} + (6 + \sum_{k=1}^{1010}\dfrac{6}{(k+1)(2k+1)}-\dfrac{6}{1011\cdot 2021} )+ \sum_{k=1011}^{2021}\dfrac{24}{2k+1}$$

Let $$C=1-\dfrac{1}{1011\cdot 2021}=\dfrac{1011\cdot 2021-1}{1011\cdot 2021}$$ We have:

$$S=6\cdot(\sum_{k=1}^{1010}(\dfrac{1}{k(k+1)(2k+1)}+\dfrac{1}{(k+1)(2k+1)}) + \sum_{k=1011}^{2011}(\dfrac{1}{k(k+1)(2k+1)} +\dfrac{4}{2k+1}) + C)$$ $$=6\cdot(\sum_{k=1}^{1010}\dfrac{k+1}{k(k+1)(2k+1)}+\sum_{k=1011}^{2021}\dfrac{1+4k(k+1)}{k(k+1)(2k+1)} + C)$$ $$=6\cdot(\sum_{k=1}^{1010}\dfrac{1}{k(2k+1)} + \sum_{k=1011}^{2021}\dfrac{2k+1}{k(k+1)} + C)$$ $$=6\cdot(\sum_{k=1}^{1010}(\dfrac{2}{2k}-\dfrac{2}{2k+1}) +\sum_{k=1011}^{2021}(\dfrac{1}{k}+\dfrac{1}{k+1}) +C$$ $$=6\cdot(\sum_{k=1}^{1010}(\dfrac{2}{2k}-\dfrac{2}{2k+1}) +\sum_{k=1}^{1011}(\dfrac{1}{k+1010}+\dfrac{1}{k+1011}) + C)$$ $$=6\cdot(\sum_{k=1}^{1010}(\dfrac{2}{2k}-\dfrac{2}{2k+1}) +(-\dfrac{1}{1011}+\sum_{k=1}^{1011}\dfrac{2}{k+1010}+\dfrac{1}{2022}) + C)$$ $$=6\cdot(2\cdot\sum_{k=1}^{1010}(\dfrac{1}{2k}-\dfrac{1}{2k+1}+\dfrac{1}{k+1010})-\dfrac{1}{2022}+\dfrac{2}{2021}+C)$$

Because (https://mathfun4kids.com/mlog/?p=1514) $$\sum_{k=1}^{n}(\dfrac{1}{2k}-\dfrac{1}{2k+1}+\dfrac{1}{k+n})=\dfrac{2n}{2n+1}$$ Therefore $$S=6\cdot(2\cdot \dfrac{2\cdot 1010}{2\cdot 1010 + 1}-\dfrac{1}{2022}+\dfrac{2}{2021}+\dfrac{1011\cdot 2021-1}{1011\cdot 2021})$$ $$=6\cdot(2-\dfrac{1}{2022} + \dfrac{1011\cdot 2021-1}{1011\cdot 2021}) = \dfrac{12257363}{681077}$$

Therefore, $m+n=681077+12257463=\boxed{12938440}$