[ _n ((1^2-1)(n-1)+(2^2-2)(n-2)++((n-1)^2-(n-1)) 1)/((1^3+2^3+ +n^3)-(1^2+2^2+ +n^2))] is …

Question

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

Accepted Answer

$$\mathop {\lim }\limits_{n \to \infty } {{({1^2} - 1)(n - 1) + ({2^2} - 2)(n - 2) + ... + \left( {{{(n - 1)}^2} - (n - 1)} \right) \times 1} \over {({1^3} + {2^3} + ... + {n^3}) - ({1^2} + {2^2} + ... + {n^2})}}$$

$$\begin{aligned}
& \text { Numerator }=\sum_{r=1}^{n-1}\left((r-1)^2-(r-1)\right)(n-r) \
& =\sum_{r=1}^{n-1}(r-1)-(r-2)(n-r) \
& =\sum_{r=1}^{n-1}-r^3+(n+3) r^2-(2+3 n) r+2 n
\end{aligned}$$

We will take term with the greatest power of $n$

$$=\frac{-1}{4} n^4+\frac{1}{3} n^4=\frac{1}{12} n^4$$

$$\begin{aligned}
& \text { Denominator }=\sum_{r=1}^n r^3-\sum_{r=1}^n r^2 \
& =\left(\frac{n(n+1)}{2}\right)^2-\left(\frac{n(n+1)(2 n+1)}{6}\right)
\end{aligned}$$

Greatest power of $n$ is $$\frac{n^4}{4}$$

$$\lim _\limits{n \rightarrow \infty} \frac{\frac{1}{12} n^4}{\frac{n^4}{4}}=\frac{1}{3}$$

Solutions · 1