Find $\lim _{n\rightarrow \infty }\frac {na^{2}_{n}+1}{\sum ^{n}_{k=1}\left( 1+2+3+\ldots +k\right) }$ Define $\left\{ a_{n}\right\} $ is an arithmetic sequence that all terms are positive integers. If $a_{10}-a_{1}=225$, find $$\lim _{n\rightarrow \infty }\dfrac {na^{2}_{n}+1}{\sum\limits^{n}_{k=1}\left( 1+2+3+\ldots +k\right) }$$
 A: Use Faulhaber's formula for computing the sums in the denominator. We have : $$\sum_{k=1}^n\sum_{j=1}^kj=\sum_{k=1}^n\frac{k(k+1)}2=\frac12\left[\sum_{k=1}^nk^2+\sum_{k=1}^nk\right]=\frac12\left[\frac{n(n+\frac12)(n+1)}3+\frac{n(n+1)}2\right]=\\=\frac{n^3}6+O(n^2)$$ On the other hand, the arithmetic progression a has the increase ratio (?) equal to $$r=\frac{a_{10}-a_1}{10-1}=\frac{225}9=25\qquad\iff\qquad a_n=25\cdot n+a_0$$ meaning that our numerator is $n(25\cdot n+a_0)^2=25^2\cdot n^3+O(n^2)$ , and our limit becomes $$\lim_{n\to\infty}\frac{625\cdot n^3+\ldots}{n^3/6+\ldots}=6\cdot625=3750.$$
A: HINT:
If $d$ is the common difference, $a_n=a_1+(n-1)d\implies a_n-a_1=(n-1)d,$
putting $n=10,225=9d\implies d=25\implies a_n=a_1+(n-1)25=25n+a_1-25$
$$na_n^2+1=n(25n+a_1-25)^2+1=625n^3+50n^2(a_1-25)+n(a_1-25)^2+1=625n^3+O(n^2)$$
As $1+2+\cdots+k-1+k=\frac{k^2+k}2$
$$\sum_{1\le k\le n}(1+2+\cdots+k-1+k)=\frac{\sum_{1\le k\le n}(k^2+k)}2=\frac{\sum_{1\le k\le n}k^2+\sum_{1\le k\le n}k}2$$
$$=\frac{\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2}2=\frac{n^3}6+O(n^2)$$
