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SOLUTION TO PROBLEM 627 PROPOSED BY GEORGE MACKIW
 

Summary: SOLUTION TO PROBLEM 627
PROPOSED BY GEORGE MACKIW
Tewodros Amdeberhan
DeVry Institute, Mathematics
630 US Highway One, North Brunswick, NJ 08902
amdberhan@admin.nj.devry.edu
Problem 627: Let x be a positive integer congruent to 1 mod 4. If n is any positive integer,
show that
1
bn,1=2cX
k=0

n
2k + 1

xk
is divisible by 2n,1
.
Solution: Using the binomial expansion of x ,1 + 1k and switching summations, we have
bn,1=2cX

  

Source: Amdeberhan, Tewodros - Department of Mathematics, Tulane University

 

Collections: Mathematics