How to obtain the quadrupole expression from the dipole expression?

I need to find the quadrupole expression from the dipole expression. I know that the dipole expression is
$$V(vec{r}) = frac{1}{4piepsilon_0}frac{vec{p}cdot (vec{r} – vec{r}^prime)}{|vec{r} – vec{r}^prime|^3} $$

Now, I wanna find the quadrupole expression from the above expression. I tried to do the following:

You can see from my diagram that I can use the dipole expression to calulate the potencial at the point as follows:
$$V(vec{r}) = frac{1}{4piepsilon_0}left[frac{-vec{p}cdot (vec{r} – vec{r}^prime)}{|vec{r} – vec{r}^prime|^3} + frac{vec{p}cdot (vec{r} – (vec{r}^prime + vec{t})}{|vec{r} – (vec{r}^prime + vec{t})|^3}right] $$

I expanded $frac{1}{|vec{r} – (vec{r}^prime + vec{t})|^3}$ into series and got
$$ frac{1}{|vec{r} – (vec{r}^prime + vec{t})|^3} = frac{1}{|vec{r} – vec{r}^prime|^3} + frac{3}{2}frac{1}{|vec{r} – vec{r}^prime|^5} left(|vec{t}|^2 – 2(vec{r}-vec{r}^prime)cdot vec{t}right) + cdots $$

Substituting in the expression of the potential

$$begin{equation}
begin{split}
V(vec{r}) &= frac{1}{4piepsilon_0}left[frac{-vec{p}cdot (vec{r} – vec{r}^prime)}{|vec{r} – vec{r}^prime|^3} + vec{p}cdot (vec{r} – (vec{r}^prime + vec{t}) left[frac{1}{|vec{r} – vec{r}^prime|^3} + frac{3}{2}frac{1}{|vec{r} – vec{r}^prime|^5} left(|vec{t}|^2 – 2(vec{r}-vec{r}^prime)cdot vec{t}right) + cdotsright] right]
&= frac{1}{4piepsilon_0}left[frac{-3[vec{p}cdot (vec{r} – vec{r}^prime)][(vec{r} – vec{r}^prime)cdot vec{t}]}{|vec{r} – vec{r}^prime|^5} + frac{vec{p}cdotvec{t}}{|vec{r} – vec{r}^prime|^3} + frac{3[(vec{r} – vec{r}^prime)cdot vec{t}][vec{p}cdotvec{t}]}{|vec{r} – vec{r}^prime|^5} + cdots right] quad text{becasue } vec{t}rightarrow 0
end{split}
end{equation}
$$

And after that I don’t know how to proceed, I need to show that the quadrupole expression is given by
$$
V(vec{r}) = frac{1}{4piepsilon_0}frac{1}{2}sum_{i,j=0}^3frac{x_ix_j}{r^5}Q_{ij}
$$
when $vec{r}^prime = 0$ and $vec{t}rightarrow 0$.

Any suggestion will be very helpful.

Thanks in advance.