Wikipedia:Reference desk/Archives/Mathematics/2011 August 2

= August 2 =

Sum formula
Good evening. The sum $$\sum_{k=0}^{n}2^k = 2^{n+1}-1$$; similarly $$\sum_{k=0}^{n}3^k = \frac{3^{n+1}-1}{2}$$, and from there induction can be used to generalise $$\sum_{k=0}^{n}q^k$$. While the formula for q=2 is fairly simple to prove, I had to use bases for the formulas for 3 and above, like so (considering 3, the argument is practically the same for any q>3): $$(\sum_{k=0}^{n}3^k)_{10}=111...(n+1,1's)...1_3$$. I then multiply both sides by 2 to get $$2(\sum_{k=0}^{n}3^k)_{10}=222...(n+1,2's)...2_3$$ (as a side question, what should I say if asked how I know 2(111...111)_3=222...222_3? it's pretty trivial but I might be asked and 'definition of multiplication' might not be satisfactory to my audience) then I just add 1 to both sides, turn the right back to decimal, and solve algebraically for $$\sum_{k=0}^{n}3^k$$. My question is, is the proof better with or without bases? The steps are the same but it's easier to see, for me at least, with bases and the analogue to 999...9+1. Thanks. (PS: I'll be presenting this to a class because my grandmother is a math teacher and she invited me to, but I don't know how familiar the class are with bases) 93.94.234.152 (talk) 00:22, 2 August 2011 (UTC)
 * How would you write the general formula in terms of $$q$$? 64.134.228.55 (talk) 01:39, 2 August 2011 (UTC)


 * In a proof by induction, it should not be necessary to prove the result for q = 3. A proof by induction usually requires a step showing that if the result is true for a given value of q then it is necessarily true for "q+1".  That is, you need to prove that assuming the result for $$\sum_{k=0}^{n}q^k$$ implies the result for $$\sum_{k=0}^{n}(q+1)^k$$. I think I would avoid using numbers in base q if possible.  As mentioned by 64.134.228.55 above, you have not stated what general result you are trying to prove.   D b f i r s   07:05, 2 August 2011 (UTC)
 * This brings me back to the time, in a distant past, when I discovered (or just proved?) the formula for a geometric sum by considering bases. But treating this rigorously, there is no a priori assurance that bases "work", and in fact development of a theory of bases would probably make heavy use of the sum formula itself. So it's best not to use bases in a "proof", but if the audience is familiar with them it will be illustrative to discuss bases and see how it all fits together.
 * For every q you can show by induction on n that the formula holds. I don't know of a way to show it by induction on q. In any case, a simpler proof than induction is to show that $$Sq-S$$ telescopes and is equal to $$q^{n+1}-1$$. -- Meni Rosenfeld (talk) 07:17, 2 August 2011 (UTC)
 * Agree with Meni there. There's not much point doing it for q an integer your way though it does give assurance the formula might be right. You'd still have to then do it for real numbers like q=1.234 for instance. Dmcq (talk) 07:21, 2 August 2011 (UTC)
 * On the other hand, I assume your audience will be familiar with base 10, so it might help to lead in with an example 111...1 = (10k-1)/9. To Dmcq, there is a theory of fractional bases which works when q is any real number greater than 1. It gets interesting when q is a Pisot–Vijayaraghavan number, see for example Golden ratio base.
 * Thanks, the Pisot–Vijayaraghavan problem sounds interesting Dmcq (talk) 08:20, 2 August 2011 (UTC)
 * Although Meni Rosenfeld already explained this proof, there happens to be an article (of course) which goes into a little more detail: Geometric_progression. Rckrone (talk) 01:54, 4 August 2011 (UTC)