Question

Consider the following sequence defined recursively: a0 = 2, a1 = 5, an = 5an-1 - 6an-2 for n > 1. Proof that an = 2^n + 3^n.

a) Mathematical induction is used to prove it.
b) The sequence is a Fibonacci sequence.
c) This sequence is not defined by a recursive formula.
d) It is impossible to prove the given formula.

Answer 1

a) Mathematical induction is used to prove it.

Step-by-step explanation:

To prove
`\(a_n = 2^n + 3^n\)` for all non-negative integers
`\(n\)` using mathematical induction, we first establish the base cases. For
`\(n = 0\), \(a_0 = 2^0 + 3^0 = 2\),` and for
`\(n = 1\), \(a_1 = 2^1 + 3^1 = 5\),` matching the given initial values. Now, assume the formula holds for
`\(k\) and \(k+1\)` for some positive integer
`\(k\).`Thus,
`\(a_k = 2^k + 3^k\) and \(a_(k+1) = 2^(k+1) + 3^(k+1)\)`. Now, we can use the recursive formula
`\(a_(n) = 5a_(n-1) - 6a_(n-2)\)` to establish
`\(a_(k+2) = 5(2^(k+1) + 3^(k+1)) - 6(2^k + 3^k)\).` By manipulating this expression, we can confirm that it simplifies to
`\(2^(k+2) + 3^(k+2)\)`. Hence, by mathematical induction, the formula
`\(a_n = 2^n + 3^n\)` is proven for all non-negative integers
`\(n\).`

Understanding mathematical induction and the recursive definition of the sequence is essential for this proof. By showing that the formula holds for the base cases and then proving the induction step, we establish the validity of the formula for all non-negative integers. The use of mathematical induction provides a rigorous and systematic approach to proving such recursive relationships.

Therefore, option (a) is the correct answer, as mathematical induction is indeed used to prove the given formula for the recursive sequence \(a_n\).