Lesson Using sequences to solve more advanced math problems

A sequence is a (finite or infinite) set of numbers such as
1, 2, 3, 4, ... or 96, 48, 24, ... that follows some sort of recursion or formula. In this lesson we will solve different math problems involving sequences.

Example 1:
Find five consecutive integers that sum up to 405.
Solution:
Instead of implementing a guess-and-check method, suppose the five integers were x, x+1, x+2, x+3, and x+4. Their sum is 405, so we have
x + x+1 + x+2 + x+3 + x+4 = 405
5x + 10 = 405
x = 79
The five integers are 79, 80, 81, 82, and 83.
This example involved a purely algebraic solution. The next example uses a sequence to prove a well-known theorem.

Example 2:
Prove that the sum of the first n odd positive integers is .
Solution: Proceed by mathematical induction. We know the case is true for n = 1, n = 2 , as , . Now suppose the statement is true for some n = k, i.e. 1+3+5+...+(2k-1) = k^2. We wish to prove that it holds for n = k+1.
The sum of the first k+1 odd positive integers is 1+3+5+...+(2k-1)+(2k+1). We have claimed that 1+3+5+...+(2k-1) = k^2 so this sum becomes k^2 + (2k+1) = (k+1)^2. Thus, if the theorem holds for some k, it will hold for k+1, and the induction is complete.

Example 3:
The Fibonacci sequence is defined a_1 = a_2 = 1, and a_k = a_(k-1) + a_(k-2). In other words, this is the sequence 1, 1, 2, 3, 5, 8, 13, etc. Prove that every fifth number in the sequence is divisible by 5.
Solution: One surefire way is to write the sequence modulo 5 and find iterative sequences. Modulo 5, the sequence becomes
1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, 0
This shows that a_21 and a_22 are congruent modulo 5, and the sequence repeats every 20 terms. As a_5, a_10, a_15, and a_20 are all 0 modulo 5, then all other a_5n will be congruent to 0 mod 5 (i.e. divisible by 5).

Example 4:
Derive a formula for the sum of an infinite geometric sequence (assume that the sequence converges)
Let . Multiplying both sides by r, we have
. The left hand side is equivalent to , or , so we have
x-1 = rx
x - rx = 1
x(1 - r) = 1
x = 1/(1 - r).