Download February 2010 Problem of the Month Solution

Okanagan College
Math Problem of the Month
FEBRUARY 2010 – Solutions
MAIN PROBLEM: Find one possible value for the next term in this sequence, and
explain the pattern: 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, ...
Solution: 1705. The sequence is similar to the Fibonacci sequence in that each number
is the sum of the previous numbers, but in this sequence it is the sum of the three
previous numbers. The next term is the sum of the last three terms given, 274 + 504 +
927, or 1705.
BONUS PROBLEM: Consider the sequences
1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...
1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, …
a) Predict the next two elements in each sequence.
b) Find a formula applying to both sequences that gives the nth element in each sequence
in terms of one or more earlier elements in the sequence.
Solution:
a) The next two elements in the first sequence are 6 and 6, while 17 and 19 are the next
two elements in the second sequence.
b) The formula for both sequences is given by: tn3  tn2  tn1  tn . In words, we add up
the two previous values in the sequence and subtract the third previous one from their
sum. The differences between the sequences are the values with which you start:
In sequence one, 1, 1, 2 are the first three terms, while 1, 2, 4 are the first three terms
in the second one.