MathFun4Kids

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

Solution: The formula is $$ S(n)=\sum_{k=1}^{n}(\dfrac{1}{2k}-\dfrac{1}{2k+1}+\dfrac{1}{k+n})=\dfrac{2n}{2n+1}$$

Proof by induction: For $n=1$, $$S(1)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{2}{3}=\dfrac{2\cdot 1}{2\cdot 1+1}$$ and the claim is correct.

Assume for $n=m$, the claim is correct, therefore $$S(m)=\dfrac{2m}{2m+1}$$ Then $$ S(m+1)=S(m)-\dfrac{1}{m+1}+(\dfrac{1}{2(m+1)}-\dfrac{1}{2(m+1)+1}+\dfrac{1}{2(m+1)})+\dfrac{1}{2m+1}$$ $$ = S(m)-\dfrac{1}{2m+3}+\dfrac{1}{2m+1}=\dfrac{2m}{2m+1}+\dfrac{2}{(2m+1)(2m+3)}$$ $$ = \dfrac{2m(2m+3)+2}{(2m+1)(2m+3)}=\dfrac{2(m+1)(2m+1)}{(2m+1)(2m+3)}=\dfrac{2(m+1)}{2(m+1)+1}$$

Therefore, for $n=m+1$, the claim is correct, and the formula is proved by induction.

Proof by expansion: By expanding the summary, it can be shown that all fractions are cancelled out, except the following $$ \dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+…+\dfrac{1}{2^m}+\dfrac{1}{2^m}-\dfrac{1}{2n+1}=1-\dfrac{1}{2n+1}=\dfrac{2n}{2n+1}$$ where $2^m$ is the largest power of 2 less or equal to $n$.