Let \(\displaystyle D(a_1,a_2,...,a_n) = \)

\(\displaystyle

det \begin{bmatrix}

a_1 + x & a_2&... & a_n\\

a_1 & a_2 +x &... & a_n\\

... & ... & ...& a_n\\

a_1 & a_2 & ...& a_n+x

\end{bmatrix} \)

and consider the system of equations :

\(\displaystyle (a_1 + x )t_1 + a_2t_2 + ... + a_nt_n = 0\)

\(\displaystyle a_1t_1 + (a_2+x)t_2 + ... + a_n t_n = 0\)

\(\displaystyle ................................................\)

\(\displaystyle a_1t_1 + a_2t_2 + ... + (a_{n-1}+x)t_{n-1} + a_nt_n = 0\)

\(\displaystyle a_1t_1 + a_2t_2 + ... + (a_n+x)t_n = 1\)

Let \(\displaystyle Y = a_1t_1 + a_2t_2 + ... + a_nt_n \) and we have

\(\displaystyle Y + xt_1 = 0 ~~...(1)\)

\(\displaystyle Y + xt_2 = 0 ~~...(2)\)

\(\displaystyle ..............\)

\(\displaystyle Y + xt_{n-1} = 0 ~~...(n-1)\)

\(\displaystyle Y + xt_n = 1 ~~...(n)\)

From \(\displaystyle (1),(2),...,(n-1) \) we obtain

\(\displaystyle t_1 = t_2 = ... = t_{n-1} = - \frac{Y}{x}\)

Consider \(\displaystyle Y = a_1t_1 + a_2t_2 + ... + a_{n-1} t_{n-1} + a_nt_n\) \(\displaystyle = (a_1 + a_2 + ... + a_{n-1} ) \left( - \frac{Y}{x} \right) + a_nt_n\) .

Hence we have these two equations with variants \(\displaystyle t_n \) and \(\displaystyle Y \)

By eliminating \(\displaystyle Y\) we obtain

\(\displaystyle t_n = \frac{ x + \sum_{i=1}^{n-1}a_i }{x(x + \sum_{i=1}^n a_i )}\)

Also by cramer's rule

\(\displaystyle t_n = \frac{ D(a_1,a_2,...,a_{n-1})}{ D(a_1,a_2,...,a_n)}\)

Therefore , \(\displaystyle \frac{ x + \sum_{i=1}^{n-1}a_i }{x(x + \sum_{i=1}^n a_i )} = \frac{ D(a_1,a_2,...,a_{n-1})}{ D(a_1,a_2,...,a_n)}\)

We have \(\displaystyle D(a_1) = x+ a_1 \) then \(\displaystyle D(a_1,a_2) = \frac{x(x+a_1+a_2)}{x+ a_1 } (x+a_1 ) = x(x+ a_1 + a_2)\) so we deduce that

\(\displaystyle D(a_1,a_2,...,a_n) = x^{n-1} ( x + \sum_{i=1}^n a_i ) \)