An arithmetic progression consists of integers. The sum of the first $n$ terms of this progression is a power of two.
Prove that $n$ is also a power of two.
Source :http://www.math.ucla.edu/~radko/circles/lib/data/Handout-967-1026.pdf
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1Hint: if you know $a_1$ and $a_n$, what is the sum of the first $n$ terms? – lulu Feb 13 '17 at 12:52
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3@Amar30657 it would be better for you if you could show your approach ! – BAYMAX Feb 13 '17 at 12:52
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Given first term $a$ and common difference $d$, sum of first $n$ terms, $S_n$ is given by $$S_n=\frac{n}{2}\left(2a+(n-1)d\right)$$ Let $S_n=2^k$ where $k\in\mathbb Z$. We must have $k\ge 0$ as the arithmetic progression consists of integers. $$2^k=\frac{n}{2}\left(2a+(n-1)d\right)$$ $$2^{k+1}=n\left(2a+(n-1)d\right)\in\mathbb Z^+$$
As the prime factorization of an integer is unique, $n$ must be of the form $2^r$, where $r\ge0$. Thus, $n$ must be a power of $2$.
GoodDeeds
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The sum of an arithmetic progression equals the number of terms times an integer or a half-integer. If twice the sum is a power of two, then both factors are a power of two.
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See that the sum of first $n$ terms is $s = \frac{n * (a_{1} + a_{n})}{2} = 2^{k}(say)$ for $k \in \mathbb{Z}$ ,now both $a_{1},a_{n}$ are integers means that $n$ must be a power of $2$.
BAYMAX
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In fact we have proved the stronger statement that if $a_i$ are the terms of any arithmetic progression, and if (for a particular $n$) $a_1+a_n$ happens to be an integer, then if the sum $s=a_1+a_2+\ldots+a_n$ is a power of two, then $n$ is a power of two. – Jeppe Stig Nielsen Feb 13 '17 at 14:29
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