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