Monthly Archives: May 2021

Given that the value of $$\sum_{k=1}^{2021}\dfrac{1}{1^2+2^2+3^2+…+k^2} + \sum_{k=1}^{1010}\dfrac{6}{2k^2-k} + \sum_{k=1011}^{2021}\dfrac{24}{2k+1}$$ can be expressed as $\dfrac{n}{m}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$. Click here for the solution. 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)} … Continue reading →

Find the formula for the following summary: $$\sum_{k=1}^{n}(\dfrac{1}{2k}-\dfrac{1}{2k+1}+\dfrac{1}{k+n})$$ Click here for the solution. 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 … Continue reading →