Problem:
There exists a unique strictly increasing sequence of nonnegative integers such that
What is ?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
First, substitute with . Then, the given equation becomes by sum of powers factorization. Now consider only . This equals . Note that equals , by difference of powers factorization (or by considering the expansion of ). Thus, we can see that forms the sum of different powers of . Applying the same method to each of , we can see that each of the pairs forms the sum of different powers of . This gives us . But we must count also the term. Thus, Our answer is .
OR
Multiply both sides by to get
Notice that , since there is a on the LHS. However, now we have an extra term of on the right from . To cancel it, we let . The two 's now combine into a term of , so we let . And so on, until we get to . Now everything we don't want telescopes into . We already have that term since we let . Everything from now on will automatically telescope to . So we let be .
As you can see, we will have to add 's at a time, then "wait" for the sum to automatically telescope for the next numbers, etc, until we get to . We only need to add 's between odd multiples of and even multiples. The largest even multiple of below is , so we will have to add a total of 's. However, we must not forget we let at the beginning, so our answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions