Problem:
Consider the sequence defined by ak​=k2+k1​ for k≥1. Given that am​+am+1​+ ⋯+an−1​=1/29, for positive integers m and n with m<n, find m+n.
Solution:
Because
k2+k1​=k(k+1)1​=k1​−k+11​
the series telescopes, that is,
1/29​=am​+am+1​+⋯+an−1​=(m1​−m+11​)+(m+11​−m+21​)+…+(n−11​−n1​)​
so (1/m)−(1/n)=1/29. Since neither m nor n is 0, this is equivalent to mn+ 29m−29n=0, from which we obtain (m−29)(n+29)=−292, or (29−m)(29+n)= 292. Since 29 is a prime and 29+n>29−m, it follows that 29−m=1 and 29+n=292. Thus m=28,n=292−29, and m+n=292−1=30⋅28=840​.
The problems on this page are the property of the MAA's American Mathematics Competitions