<s>[INST] 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 $$\lim _{n\rightarrow \infty }\dfrac {na^{2}_{n}+1}{\sum\limits^{n}_{k=1}\left( 1+2+3+\ldots +k\right) }$$
 [/INST] 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 $1+2+\cdots+k-1+k=\frac{k^2+k}2$
$$\sum_{1\le k\le n}(1+2+\cdots+k-1+k)=\frac{\sum_{1\le k\le n}(k^2+k)}2=\frac{\sum_{1\le k\le n}k^2+\sum_{1\le k\le n}k}2$$
$$=\frac{\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2}2=\frac{n^3}6+O(n^2)$$
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