Sum-equation!



  • Solve in N\mathbb{N} the following equation:
    k=1k=nk(2(2k2+1)+4k4+1)2k2+2k+1+2k22k+1=18.\sum_{k=1}^{k=n}\frac{k\left (2(2k^2+1)+\sqrt{4k^4+1} \right )}{\sqrt{2k^2+2k+1}+\sqrt{2k^2-2k+1}}=\frac{1}{8}.


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