The fraction and representation of powers of $2$ Let $a_1<a_2<...<a_k$ be an increasing sequence of nonnegative integers such that $$\frac{2^{289}+1}{2^{17}+1}=2^{a_1}+2^{a_2}+...+2^{a_k}$$ Then find the value of k.
My approach : Since $$\frac{2^{289}+1}{2^{17}+1}=2^{255}(2^{16}+2^{15}+...+2+1)+2^{221}(2^{16}+2^{15}+...+2+1)+...+$$ $$2^{51}(2^{16}+2^{15}+...+2+1)+2^{17}(2^{16}+2^{15}+...+2+1)+1$$ I observed that $k=271$. But the answers (in multiple choice) are $117,136,137,273,306$. Did I miss something???
Note : I think since the powers are in increasing order this representation is unique. The sum of powers of $2$ (which are >1) cannot be power of $2$. 
What is the wrong in my solution?
 A: It turned out that the OP did the difficult part of finding the binary expansion correctly, but then simply miscounted or misread the question mistaking $k$ for $a_k$ (the latter is the OP's $271$). There are 8 groups of 17 powers of two and that lonely $2^0$ giving
$$8\cdot17+1=137$$ as the answer.

To the reader who did not see where the expansion came from:
We can use the polynomial factorization (think of it as a geometric sum with ratio $-x$)
$$
\frac{x^{17}+1}{x+1}=(x^{16}-x^{15})+(x^{14}-x^{13})+\cdots+(x^2-x)+1. \qquad(*)
$$
Plug in $x=2^{17}$. Each group in parens becomes
$$
(x^{2k+2}-x^{2k+1})=x^{2k+1}(x-1)=2^{17(2k+1)}(2^{17}-1), k=0,1,2,\ldots,7.
$$
Here $2^{17}-1=\sum_{j=0}^{16}2^j$, another geometric sum. The OPs expansion follows by replacing each binomial term (in parens) in $(*)$ with this.
A: To avoid confusion, let $2^{17}=a$, considering $289=17^2$, we have:
$\frac{a^{17}+1}{a+1}=a^{16}-a^{15}+a^{14}-a^{13}+ . . +1$
Therefor:
$a_k=(17).(16)=272$
$a_{k-1}=(17).(15)=255$
$a_{k-2}=(17).(14)=238$
.
.
.
$a_2=(17).1=17$
$a_1=0$
Now number of terms in bracket is from 0 to 272, therefore $k=272+1=273$.Note that the terms in bracket are in fact reduced form, because intermediate terms are cancelled two by two. If you divide numerator by denominator.The degree of quotient is $289-17=272$ and number of its terms is $273$.
