Problem:
Two distinct, real, infinite geometric series each have a sum of 1 and have the same second term. The third term of one of the series is 1/8, and the second term of both series can be written in the form pmββnβ, where m,n, and p are positive integers and m is not divisible by the square of any prime. Find 100m+10n+p.
Solution:
Let the two series be
k=0βββaβ rk and k=0βββbβ sk
The given conditions imply that a=1βr,b=1βs, and ar=bs. It follows that r(1βr)=s(1βs), that is rβs=r2βs2. Because the series are not identical, rξ =s, leaving r=1βs as the only possibility, and the series may be written as
k=0βββ(1βr)β rk and k=0βββrβ (1βr)k
As we may pick either series as the one whose third term is 1/8, set (1βr)r2=1/8, from which we obtain 8r3β8r2+1=0. The substitution t=2r yields t3β2t2+1=0, for which 1 is a root. Factoring gives (tβ1)(t2βtβ1)=0, so the other two roots are (1Β±5β)/2, which implies that r=1/2 or r=(1Β±5β)/4. However, if r were 1/2, the two series would be equal; and if r were (1β5β)/4, then s would be (3+5β)/4, but convergence requires that β£sβ£<1. Thus r=(1+5β)/4, and β1<s=(3β5β)/4<1. The second term of the series is therefore equal to