Question

produs de la k=2 la n de k^3+1/k^3-1

Answer 1

$\displaystyle\bf\\\prod_{k=2}^n\frac{k^3+1}{k^3-1} =\prod_{k=2}^n \frac{(k+1)(k^2-k+1)}{(k-1)(k^2+k+1)} = \prod_{k=1}^n \frac{k+1}{k-1} \frac{k^2-k+1}{k^2+k+1} = \\\\\frac{\prod_{k=2}^n k+1 }{\prod_{k=2}^n k-1} \frac{\prod_{k=2}^nk^2-k+1}{\prod_{k=2}^nk^2+k+1} =\frac{\prod_{k=2}^n k+1 }{\prod_{k=2}^n k-1} \frac{ \prod_{k=2}^n (k-1)^2+(k-1)+1}{\prod_{k=2}^n k^2+k+1}=$

$\displaystyle\bf\\\frac{\prod_{k=3}^{n+1}k}{\prod_{k=1}^{n-1}k} \frac{\prod_{k=1}^{n-1}k^2+k+1}{\prod_{k=2}^{n}k^2+k+1} = \frac{n(n+1)}{2} \frac{3}{n^2+n+1} = \frac{3}{2} \frac{n(n+1)}{n^2+n+1}$