Show that $\frac{1\times 2^2+2\times 3^2+\dot{\dot{\dot{+}}}n\times {\left(n+1\right)}^2}{1^2\times 2+2^2\times 3+\dot{\dot{\dot{+}}}{n}^2\times \left(n+1\right)}=\frac{3n+5}{3n+1}\dot{}$

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