My Dad's Favorite

We are going to find a polynomial with roots $t=\dfrac{1}{1-x}$ , so solving for $x$ :

$x=\dfrac{t-1}{t}$

The equation is $\dfrac{x^{151}-1}{x-1}=0$ for $x\neq 1$ , so $t\neq 0$ :

$\dfrac{\left(\dfrac{t-1}{t}\right)^{151}-1}{\dfrac{t-1}{t}-1}=0$

$t^{151}-(t-1)^{151}=0$

If we expand it using the binomial theorem, we get:

$\dbinom{151}{1}t^{150}-\dbinom{151}{2}t^{149}+\dbinom{151}{3}t^{148}+\cdots=0$

By Vieta's formulas, the sum we want is $\dfrac{\text{coefficient of }t^{148}}{\text{coefficient of }t^{150}}$ :

$\dfrac{\dbinom{151}{3}}{\dbinom{151}{1}}=\dfrac{\dfrac{151\times150\times149}{2\times 3}}{151}$

$\dfrac{150 \times 149}{6}=\boxed{3725}$

In general, for the equation $x^n+x^{n-1}+\cdots+x+1=0$ , the sum is $\dfrac{n(n-1)}{6}$

Since $x_{1}$ , $x_{2}$ , $x_{3}$ $\ldots$ $x_{150}$ are the roots $f(x)=0$ . It implies that $f(x) = (x-x_{1})(x-x_{2})\ldots (x-x_{150})$ .

It's easily observable that by differentiating $f(x)$ twice w.r.t $x$ . We get,

$f''(x) = 2f(x) \left(\sum_{1 \leq i < j \leq 150} \dfrac{1}{(x-x_{i})(x-x_{j})} \right)$

So, $\sum_{1 \leq i < j \leq 150} \dfrac{1}{(1-x_{i})(1-x_{j})} = \dfrac{f''(1)}{2f(1)}$

But, $f(1) = \underbrace {1+1+\ldots + 1} _{151 \text{number of 1's}} = 151$

And $f''(x) = 150\times 149 x^{148} +149 \times 148 x^{147} + \ldots + 2 \times 1$ . So, $f''(1) = 150 \times 149 + 149 \times 148 \ldots + 2 \times 1 = 1124950$

So, therefore

$\sum_{1 \leq i < j \leq 150} \dfrac{1}{(1-x_{i})(1-x_{j})} = \dfrac{1124950}{2 \times 151} =\boxed{3725}$

Note that $\prod_{k=1}^{150}{(1-x_k)}=f(1)=151$ , so that the sum we seek is $\frac{1}{151}\sum_{i<j}\prod_{k\neq{i,j}}(1-x_k)$ . If we consider the polynomial $g(x)=x^{151}-1$ , with the additional root $x_0=1$ , then the sum we seek is $\frac{1}{151}\sum_{i<j<p}\prod_{k\neq{i,j,p}}(1-x_k)$ , a symmetric polynomial of degree 148 in the 151st roots of unity. The constant term of this polynomial is $\frac{1}{151}\dbinom{151}{3}=\boxed{3725}$ , and all the elementary symmetric polynomials of degree 1 through 150 vanish, by Viete.

the sum we want é $S= \dfrac{\binom{150}{148}+\binom{149}{147}+\cdots+\binom{2}{0}}{151}$

or $S= \dfrac{ \binom{151}{148}}{151}= 3725$ .