How to show $\lim_{n\to\infty}\sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4$? How would I show $$\lim_{n\to\infty}\sum_{k=1}^n(1+(k)/n)^3(1/n)=15/4?$$

My attempt is using the Riemann Sum technique. We know $(1+(k)/n)^2=f(\zeta_k)$ and $(1/n)=\Delta x$. So the definite integral goes from $a$ to $b$ where $b-a=1$.

I think I should find the formula for $$\sum_{k=1}^nk^3.$$ This would be $1+8+27+\cdots+n^3$... but then I get stuck.
Hint:

$$S_n:=\sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.

Now

$$S_n-S_{n-1}=an^4+bn^3+cn^2+\cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-\cdots
\\=4an^3+bn^3-bn^3+\cdots$$ and lower order terms. You conclude that

$$S_n\sim\frac{n^4}4$$ and this is enough for you to evaluate the limit.
Hint:

$$S_n:=\sum_{k=1}^n k^3$$ must be a quartic polynomial in $n$ because $S_{n}-S_{n-1}=n^3$ is a cubic polynomial.

Now

$$S_n-S_{n-1}=an^4+bn^3+cn^2+\cdots-a(n-1)^4-b(n-1)^3-c(n-1)^2-\cdots
\\=4an^3+bn^3-bn^3+\cdots$$ and lower order terms. You conclude that

$$S_n\sim\frac{n^4}4$$ and this is enough for you to evaluate the limit.