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limn→∞12(n)+22(n−1)+32(n−2)+⋯+n2(1)13+23+33+⋯+n3\large \displaystyle\lim_{n\to\infty}\dfrac{1^2(n)+2^2(n-1)+3^2(n-2)+\cdots+n^2(1)}{1^3+2^3+3^3+\cdots +n^3} n→∞lim13+23+33+⋯+n312(n)+22(n−1)+32(n−2)+⋯+n2(1)
Evaluate the limit to 3 decimal places.
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