n lim(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 equal…

Question

$\underset{n	o \infty}{lim}\frac{({1}^{2}-1)(n-1)+({2}^{2}-2)(n-2)+\cdots +((n-1{)}^{2}-(n-1))\cdot 1}{({1}^{3}+{2}^{3}+\cdots \cdots +{n}^{3})-({1}^{2}+{2}^{2}+\cdots \cdots +{n}^{2})}$ is equal to :

Accepted Answer

$\underset{n	o \infty}{lim}\frac{({1}^{2}-1)(n-1)+({2}^{2}-2)(n-2)+...+({(n-1)}^{2}-(n-1))	imes 1}{({1}^{3}+{2}^{3}+...+{n}^{3})-({1}^{2}+{2}^{2}+...+{n}^{2})}$

$\begin{gathered} =\sum _{r=1}^{n-1}((r-1{)}^{2}-(r-1))(n-r) \ =\sum _{r=1}^{n-1}(r-1)-(r-2)(n-r) \ =\sum _{r=1}^{n-1}-{r}^{3}+(n+3){r}^{2}-(2+3n)r+2n \end{gathered}$

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

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

$\begin{gathered} =\sum _{r=1}^{n}{r}^{3}-\sum _{r=1}^{n}{r}^{2} \ ={(\frac{n(n+1)}{2})}^{2}-(\frac{n(n+1)(2n+1)}{6}) \end{gathered}$

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

$\underset{n	o \infty}{lim}\frac{\frac{1}{12}{n}^{4}}{\frac{{n}^{4}}{4}}=\frac{1}{3}$

Solutions · 1