Proof by induction, puzzles by answer

  • Thread starter Thread starter MegaDeth
  • Start date Start date
  • Tags Tags
    Induction Proof
MegaDeth
Messages
83
Reaction score
0
Ok, so there's this proof by induction question. I looked at the answer for it but I don't understand it.

TuXle.jpg


How does it get from being 2^(k+1) to being 2^(k+2)?
 
Last edited by a moderator:
Physics news on Phys.org
You need to provide a little more background into your question, I think.
 
Your image did not link properly so it is impossible to answer your question.
 
MegaDeth said:
How does it get from being 2^(k+1) to being 2^(k+2)?
2^{k+1}(k-1+k+1) = 2^{k+1}(2k) = 2^{k+2}k
 
you got [2^(k+1)](2k).

multiply 2^(k+1) by two, that is raising your exponent by 1. Hence 2^(k+2)
 
Last edited:
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1 Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers. So we only handle with original Zeta...
Back
Top