\lim \limits_{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)} is equal to :

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Answer: D. \frac{1}{3} \mathop {\lim }\limits_{n 	o \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} ight) 	imes 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}} \begin{aligned} & 	ext { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)ight)(n-r) \ & =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \ & =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n \end{aligned} We will take term with the greatest power of n =\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4 \begin{aligned} & 	ext { Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \ & =\left(\frac{n(n+1)}{2}ight)^2-\left(\frac{n(n+1)(2 n+1)}{6}ight) \end{aligned} Greatest power of n is \frac{n^4}{4} \lim \limits_{n ightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}

$$\lim \limits_{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)}$$ is equal to :

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