*if it is in parentheses, it is subscripted.
the sequence {a(n)} is defined recursively by a(1) = 1, a(2) = 1 and for all n>= 1, a(n+2) = a(n+1) + a(n). use the principle of mathematical induction to prove that a(1) + a(2) + a(3) + ... + a(n) = a(n+2) - 1
To prove the given statement using the principle of mathematical induction, we need to follow a three-step process:
1. Base case: Show that the statement holds for the base case n = 1.
2. Inductive hypothesis: Assume that the statement holds for some arbitrary positive integer k, i.e., a(1) + a(2) + a(3) + ... + a(k) = a(k+2) - 1.
3. Inductive step: Prove that the statement also holds for the next integer, k + 1.
Let's go through each step in detail:
1. Base case (n = 1):
Substituting n = 1 in the given equation, we have:
a(1) = a(1+2) - 1
a(1) = a(3) - 1
From the recursive definition of the sequence, we know that a(3) = a(2) + a(1). Since a(2) = 1 and a(1) = 1, we can substitute these values:
a(1) = 1+1 - 1
a(1) = 2 - 1
a(1) = 1
Hence, the statement holds true for the base case.
2. Inductive hypothesis (assume true for k):
Assume that the statement holds for some positive integer k, i.e., a(1) + a(2) + a(3) + ... + a(k) = a(k+2) - 1. This is our inductive hypothesis.
3. Inductive step (prove for k + 1):
We need to prove that the statement holds for the next positive integer, k + 1, i.e., a(1) + a(2) + a(3) + ... + a(k) + a(k+1) = a(k+3) - 1.
Using the recursive definition of the sequence, we can express a(k+1) as a(k) + a(k-1):
a(1) + a(2) + a(3) + ... + a(k) + a(k+1) = a(k+2) - 1 + (a(k) + a(k-1))
By the inductive hypothesis, a(1) + a(2) + a(3) + ... + a(k) = a(k+2) - 1. Substituting this, we get:
(a(k+2) - 1) + a(k) + a(k-1)
Expanding and rearranging terms:
(a(k+2) + a(k) + a(k-1)) - 1
From the recursive definition, we know that a(k+3) = a(k+2) + a(k+1). Substituting this, we get:
a(k+3) - 1
This shows that a(1) + a(2) + a(3) + ... + a(k) + a(k+1) = a(k+3) - 1, which proves the statement for k + 1.
By the principle of mathematical induction, we have proven that a(1) + a(2) + a(3) + ... + a(n) = a(n+2) - 1 for all positive integers n.