An Easy Way And A Hard Way I

Since we know that $x+\frac{1}{x}=1$ , lets square both sides and get

$x^{2}+2+\frac{1}{x^{2}}=1$ $x^{2}+\frac{1}{x^{2}}=-1$

For any value $x^{k}$ such that $x^{k}+\frac{1}{x^{k}}=-1$ , squaring both sides will get us

$x^{2k}+2+\frac{1}{x^{2k}}=1$ $x^{2k}+\frac{1}{x^{2k}}=-1$

Let $k=2^{n}$ for some integer value $n$ . Then we can rewrite our equation as

${x^{2^{n}}+\frac{1}{x^{2^{n}}}=-1}$

Now, squaring both sides will get us

${x^{2^{n+1}}+\frac{1}{x^{2^{n+1}}}=-1}$

This fits right into our summation, since we are looking to sum up all values of the expression

$x^{2^{j}}+\frac{1}{x^{2^{j}}}$

for integer values of $j$ between $0$ and $100$ .

We know that for $j=0$ our expression equals $1$ , because that was given. We solved for $j=1$ and got that our expression equaled $-1$ , and we know that if the expression equals $-1$ for some integer value $n$ then our expression evaluated at $n+1$ would equal $-1$ .

So for all values of $j$ between and including $1$ and $100$ , our expression equals $-1$ , and so our summation results in

$\left|1+(100)(-1)\right|$

which equals

$\large \boxed{99}$

If $\alpha$ satisfies the given constraint, then so does $\frac{1}{\alpha}$ . So the given sum can be rewritten as
$\lvert \sum_{j=0}^{100} \alpha^{2^j} + \alpha^{2^{-j}} \rvert = \lvert \sum_{j=0}^{100} \alpha^{2^j} + \beta^{2^{j}} \rvert = \lvert \sum_{j=0}^{100} V_n \rvert$
Here, $\alpha, \beta$ are the roots of equation $x^2 - x + 1$ , which is our constraint rearranged and $V_n = \alpha^n + \beta^n$ .
Using newton's sums, $V_n = V_{n-1} - V_{n-2}$
From n = 1 to 6, $V_n$ follows the pattern 1, -1, -2, -1, 1, 2 and keeps repeating it. As $2^n$ will be either of the form $6m+ 2$ or $6m+4$ for $n > 1$ , $V_{2^j} = -1 \forall j \ge 1$ and the required sum will be $\lvert 1 + 100\times (-1) \rvert = \boxed{99}$ .