Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$ To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is 
\begin{align}
\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T
\end{align}
in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In addition, $(\cdot)^T$ is transposition operation. $\mathbf{G}$ is $u\times u$. We use $\mathbf{G}_u$ to underline the dimension of $\mathbf{G}$.
The question is to determine $\det(\mathbf{G})$.
As I know: We rewrite $\mathbf{G}_u$ as 
\begin{align}
\mathbf{G}_u=\left[\begin{array}{ccc}
\mathbf{G}_{u-1} & b\\
b & a+b
\end{array}
\right]
\end{align}
Using the determinant of  block matrix lemma 
\begin{align}
\det\left[\begin{array}{ccc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}
\end{array}
\right]=\det(\mathbf{A})\det(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})
\end{align}
We then have 
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})
\end{align}
It still needs to get $\mathbf{G}_{u-1}^{-1}$ via matrix inverse lemma
\begin{align}
(\mathbf{A}+\mathbf{BC})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1}\mathbf{B}(\mathbf{I}+\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1}
\end{align}
We then have 
\begin{align}
\mathbf{G}_{u}
&=\frac{1}{a}\mathbf{I}-\frac{1}{a^2}b\boldsymbol{e}\left(1+\frac{b}{a}\boldsymbol{e}^T\boldsymbol{e}\right)^{-1}\boldsymbol{e}^T\\
&=\frac{1}{a}\mathbf{I}-\frac{b}{a(a+bu)}\boldsymbol{ee}^T
\end{align}
where $\boldsymbol{e}^T\boldsymbol{e}=u$ is used. Plugging it into 
\begin{align}
&\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})\\
=&a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)
\end{align}
Then 
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\left[{a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)}\right]
\end{align}
Although, the connection between $\mathbf{G}_u$ and $\mathbf{G}_{u-1}$ is found, I can't give the expression of $\mathbf{G}_u$. Please give me hand, thanks a lot!

A More Basic Approach
Let us consider for illustration, $G_4$:
$$
G=aI_{4\times4}+bee^T \\
\text{where } e = \begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix}
\\
\text{Hence, } G = 
  \begin{bmatrix}
   a & 0 & 0 & 0 \\
   0 & a & 0 & 0 \\
   0 & 0 & a & 0 \\
   0 & 0 & 0 & a \\ 
  \end{bmatrix} +
  \begin{bmatrix}
   b & b & b & b \\
   b & b & b & b \\
   b & b & b & b \\
   b & b & b & b \\
  \end{bmatrix} =
  \begin{bmatrix}
   a + b & b & b & b \\
   b & a + b & b & b \\
   b & b & a + b & b \\
   b & b & b & a + b \\
  \end{bmatrix}
$$
Taking the determinant, let us apply the transformation $\left(R_1 \rightarrow R_1 + R_2 + R_3 + R_4 \right)$,
$$
\det(G) = \begin{vmatrix}
   a + 4b & a + 4b & a + 4b & a + 4b \\
   b & a + b & b & b \\
   b & b & a + b & b \\
   b & b & b & a + b \\
  \end{vmatrix}
$$
Applying now $\left( R_1 \rightarrow \frac{1}{a+4b} R_1 \right)$ and $\prod_{k=2}^{4}{\left(R_k \rightarrow R_k - bR_1\right)}$
$$
= (a+4b) \cdot \begin{vmatrix}
   1 & 1 & 1 & 1 \\
   0 & a & 0 & 0 \\
   0 & 0 & a & 0 \\
   0 & 0 & 0 & a \\ 
  \end{vmatrix}
\\
\text{So } \det{(G)} = (a+4b)\cdot a^3
$$
Similarly, for $G_n$, upon applying the transformations $\left( R_1 \rightarrow \frac{1}{a+nb} \sum_{k=1}^{n}{R_k} \right)$ and $\left( \prod_{k=2}^{n} \left( R_k \rightarrow R_k - bR_1 \right) \right)$ and then simply expanding,
$$
\boxed {\det{(G_n)} = (a+nb) \cdot a^{n-1}}
$$
