Difference between revisions of "Besot Power Series"

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===Problem 2===
 
===Problem 2===
 
<math>\sum\limits_{i=1}^{2016} n^i = n^{2017}</math>. What is <math>n</math>?
 
<math>\sum\limits_{i=1}^{2016} n^i = n^{2017}</math>. What is <math>n</math>?
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Latest revision as of 21:13, 7 October 2024

Theorem

Besot's Power Series Theorem states that $\sum\limits_{i=1}^m n^{a+(i-1)z} = \frac{n^{a+mz}-n^a}{n^z-1}$

Proof

Let there be a sum $n^a + n^{a+z} + ... n^{a+(m-1)z} = \sum\limits_{i=1}^m n^{a+(i-1)z} = s$

$s = \sum\limits_{i=1}^m n^{a+(i-1)z}$

$sn^z = \sum\limits_{i=1}^m n^{a+iz}$

$sn^z-s = \sum\limits_{i=1}^m n^{a+iz} - \sum\limits_{i=1}^m n^{a+(i-1)z}$

$s(n^z-1) = n^{a+mz}-n^a$

$s = \frac{n^{a+mz}-n^a}{n^z-1}$

$\boxed{\frac{n^{a+mz}-n^a}{n^z-1}}$


Problems

Problem 1

What is $2^3+2^5+2^7?$

Problem 2

$\sum\limits_{i=1}^{2016} n^i = n^{2017}$. What is $n$?

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