This is quite straightforward. Let's start with:
M\vec{R} = \sum_i m_i \vec{r}_i
Square it:
M^2 R^2 = \sum_{i,j} m_i m_j \vec{r}_i \cdot \vec{r}_j \qquad (1)
Then consider \vec{r}_{i j} \equiv \vec{r}_i - \vec{r}_j \Longleftrightarrow r^2_{ij} = (\vec{r}_i - \vec{r}_j)^2 = r^2_i - 2 \vec{r}_i \cdot \vec{r}_j + r^2_j \Longleftrightarrow \vec{r}_i \cdot \vec{r}_j = \frac{1}{2}(r^2_i+r^2_j-r^2_{ij})
Insert the last relationship in (1) and you'll obtain:
M^2 R^2 = \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j-r^2_{ij}) = \frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) - \frac{1}{2} \sum_i m_i m_j r^2_{ij} \qquad (2)
Now, consider the first term of (2)'s rhs:
\frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) = \frac{1}{2}\sum_i m_i \left[\sum_j m_j (r^2_i + r^2_j) \right] = \frac{1}{2} \sum_i m_i \left[\sum_j m_j r^2_i + \sum_j m_j r^2_j \right] \qquad (3)
But \sum_j m_j r^2_i = M r^2_i so:
\frac{1}{2} \sum_i m_i \left[\sum_j m_j r^2_i + \sum_j m_j r^2_j \right] = \frac{1}{2} \sum_i m_i \left[ M r^2_i + \sum_j m_j r^2_j \right] = \frac{1}{2} \left[ M \sum_i m_i r^2_i + \sum_i m_i \sum_j m_j r^2_j \right] = \frac{1}{2} \left[ M \sum_i m_i r^2_i + M \sum_j m_j r^2_j \right] =
= \frac{1}{2} M \left[ \sum_i m_i r^2_i + \sum_j m_j r^2_j \right]
Because "j" is a summed index, you can call it "i" so (3) becomes:
\frac{1}{2} \sum_{i,j} m_i m_j (r^2_i+r^2_j) = \frac{1}{2} M \left[ \sum_i m_i r^2_i + \sum_i m_i r^2_i \right] = M \sum_i m_i r^2_i \qquad (3)
Finally, just insert rhs of (3) in (2) and that's it
