Problem:
The increasing geometric sequence x0,x1,x2,… consists entirely of integral powers of 3. Given that
n=0∑7log3(xn)=308 and 56≤log3(n=0∑7xn)≤57
find log3(x14).
Solution:
The sequence is geometric, so there exist numbers a and r such that xn=arn. It follows that
308=n=0∑7log3(xn)=n=0∑7log3(arn)=n=0∑7[log3(a)+nlog3(r)]=8log3(a)+(n=0∑7n)log3(r)=8log3(a)+28log3(r).
Thus 2log3(a)+7log3(r)=77. Furthermore,
log3(n=0∑7xn)=log3(n=0∑7arn)=log3(a⋅r−1r8−1)=log3(ar7⋅1−r11−r81)=log3(a)+7log3(r)+log3(1−r11−r81)
which is between 56 and 57. Because the terms are all integral powers of 3, it follows that a and r must be powers of 3. Also, the sequence is increasing, so r is at least 3. Therefore
1=1−r11−r1<1−r11−r81<1−311=23<3, and 0<log3(1−r11−r81)<1.
Also note that since a and r are powers of 3,log3(a)+7log3(r) is an integer and therefore must equal 56. Thus log3(a)+7log3(r)=56. The two equations log3(a)+7log3(r)=56 and 2log3(a)+7log3(r)=77 have the solution log3(a)= 21 and log3(r)=5. It follows that log3(x14)=log3(ar14)=log3(a)+14log3(r)=21+14⋅5=91.
The problems on this page are the property of the MAA's American Mathematics Competitions