Wonderfully Scary

The coefficient for $x^n$ is given by:

$a_n = (-1)^n \sum _{i=1} ^{102} {\prod _{j=0} ^{n} {2^j } }$

Therefore,

$a_{101} = - (2^02^12^2...2^{98}2^{99}2^{100} ) - (2^02^12^2...2^{98}2^{99}2^{101} )$

$\quad \quad \space \space - (2^02^12^2...2^{98}2^{100}2^{101}) -...- (2^12^22^3...2^{99}2^{100}2^{101})$

$\quad = - (2^02^12^2...2^{99}2^{100}2^{101}) ( \frac {1} {2^0} + \frac {1} {2^1} + \frac {1} {2^2} + ... + \frac {1} {2^{101} } )$

$\quad = - 2^{ \sum _{i=0} ^{101} {i} } \sum _{j=0} ^{101} {\frac {1}{2^j} } = - 2 ^{5151} \left( \frac {1- \frac{1}{2^{102}} }{1 - \frac{1}{2} } \right)$

$\quad = - 2 ^{5151} (2) (1 - 2^{-102} ) = - 2 ^{5152} + 2 ^{5050}$

$\quad = \boxed{2 ^{5050} - 2 ^{5152}}$

What is given to us is a polynomial with roots $1,\dfrac {1}{2}, \dfrac{1}{4}$ and so on..(Form a G.P. with common ration .5))

Now, we know sum of roots is $\dfrac{-b}{a}$ , where a and b are the terms with highest and the one with (highest -1 degree) .In this case,

$a$ is $-1^{102}. 2^1.2^2.2^3......2^{101}$

Applying $S = \dfrac{n(n+1)}{2}$ where S is the sum of first n natural no's, we get power of 2 equal to $5151$

S, we have $a = 2^{5151}$

Sum of n terms of a G.P. = $A.\dfrac{r^n-1}{r-1}$ , where A is the first term of G.P. , r is the common ratio. Plugging in the values, we get the sum of roots as $2 - 2^{-101}$

Now $\dfrac {-b}{2^{5151}} = 2-2^{-101}$

So $b = 2^{5050}-2^{5152}$