Try this problem from TOMATO Problem 7 based on the Parity of the terms of a sequence.

**Problem: Parity of the terms of a sequence**

If \( a_0 = 1 , a_1 = 1 \) and \( a_n = a_{n – 1} a_{n – 2} + 1 \) for \( n > 1 \), then:

(A) \(a_{465} \) is odd and \(a_{466} \) is even;

(B) \(a_{465} \) is odd and \(a_{466} \) is odd;

(C) \(a_{465} \) is even and \(a_{466} \) is even;

(A) \(a_{465} \) is even and \(a_{466} \) is odd;

**Discussion:**

First we note a pattern and then we prove that the pattern actually holds.

Note that:

\( a_0 = 1 \) is odd

\( a_1 = 1 \) is odd

\( a_2 = a_0 a_1 + 1 = 1\times 1 + 1 = 2 \) is even

\( a_3 = a_1 a_2 + 1 = 1 \times 2 + 1 = 3 \) is odd

\( a_4 = a_2 a_3 + 1 = 2 \times 3 + 1 = 7 \) is odd

\( a_5 = a_3 a_4 + 1 = 3 \times 7 + 1 = 22 \) is even

So the pattern that we observe is the following order: odd, odd, even, odd, odd, even…

We show this by strong form of induction. Suppose this pattern holds true for all n upto n = 3k+2

(that is \( a_{3k+2} = even , a_{3k+1} = odd, a_{3k}= odd \) ).

Our computations show that this is true for k =1 (so for initial value it is true).

Let us show for the next three values:

\( a_{3k+3} = a_{3k+2} \times a_{3k+1} + 1 = even \times odd + 1 = odd \)

\( a_{3k+4} = a_{3k+3} \times a_{3k+2} + 1 = odd \times even + 1 = odd \)

\( a_{3k+5} = a_{3k+4} \times a_{3k+3} + 1 = odd \times odd + 1 = even \)

Thus we showed that whenever the index is of the form 3j+2, the number is even, otherwise if the index is of the form 3j or 3j+1, the term is odd.

Since 465 and 466 are respectively of the form 3j and 3j+1, hence

\( a_{465} \) and \( a_{466} \) both are odd.

## Chatuspathi:

**What is this topic:**Induction**What are some of the associated concepts:**Strong form of induction**Where can learn these topics:**Cheenta I.S.I. & C.M.I. course, Math Olympiad Program discusses these topics in the ‘Induction’ module.**Book Suggestions:**Elementary number theory by David Burton

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