Using induction to show one sequence is equal to the other

Question

I am training for olympiad and came across this question in a book "recurrent sequences". The original question was:

Find the term $a_{2020}$ for the sequence $(a_{n})_{n≥1}$ defined by $a_{1} = 1$ and
$a_{n+1} = 2a_{n} + \sqrt{3a_{n}^2 − 2}$,
$n = 1, 2, \ldots$

My progress:

I calculate a few terms and eventually found the pattern that

$b_{n+1} = 4b_{n} - b_{n-1}$ gives terms that are equal to the sequence in the question. The nth term for this sequence could be calculated much more easily, so I tried to use induction to prove that $b_{n} = a_{n}$ for all $n$. I have no idea how I would do the induction for this. I have a base case with $b_{1} = a_{1}$, then I set

$a_{k} = b_{k}$ for some $k > 1$, then replaced the $a_{k} and b_{k}$ with their definitions, and if we chose k such that it is the smallest value for which this hypothesis does not hold, then $a_{k-1} = b_{k-1}$ and the same for the index $k-2$ terms that appear with the definition substituition. From here where would I go?

The book solution:

The book's solution was much nicer. The sequence was rearranged to give

$a_{n+1}^2-4a_{n}a_{n+1}+a_{n}^2+2=0$

Then the index was replaced with $n-1$ to get

$a_{n}^2-4a_{n-1}a_{n}+a_{n-1}^2+2=0$

Since both equations are quadratics with the same constant coefficients, we can say that $a_{n+1}$ and $a_{n-1}$ are the roots of the quadratic

$t^2-4a_{n}t+a_{n}^2+2 = 0$.

Then, using Vieta's formulas the relation $a_{n-1} + a_{n+1} = 4a_{n}$ can be achieved from where the explicit formula is achieved.

Answer 1

Presumably you want to show that $a_{n+1}= 4 a_n - a_{n-1} $, which can be guessed by calculating the first few terms.

If so, we need to show that $ 2a_n - a_{n-1} = \sqrt{ 3a_n^2 - 2}$, or that $2a_n - a_{n-1} >0$ (which is obvious) and $ a_n^2 - 4a_na_{n-1} + a_{n-1}^2 = -2$. The latter can be obtained by expanding out $(a_n - 2a_{n-1} )^2 = 3a_{n-1}^2 - 2$ (which was also done in the book's solution).

So we didn't need to apply Vieta's formula, and could have proven this by careful arithmetic.

Answer 2

Find the term $a_{2020}$ for the sequence $(a_{n})_{n≥1}$ defined by $a_{1} = 1$ and
$a_{n+1} = 2a_{n} + \sqrt{3a_{n}^2 − 2}$,
n = 1, 2,....

You have that $~a_1 = 1, ~a_2 = 3, ~a_3 = 11, ~a_4 = 41, ~a_5 = 153, a_6 = 571.$ So, the hypothesis that $~a_{n+1} = 4a_n - a_{n-1}~$ is correct for the first few terms.

You have that

$$a_n - 2a_{n-1} = \sqrt{3a_{n-1}^2 - 2} \implies $$

$$a_n^2 - 4a_na_{n-1} + 4a_{n-1}^2 = 3a_{n-1}^2 - 2 \implies $$

$$a_n^2 - 4a_na_{n-1} + a_{n-1}^2 + 2 = 0 \implies $$

$$3a_n^2 - 2 = 4a_n^2 - 4a_na_{n+1} + a_{n-1}^2 = \left[ ~2a_n - a_{n-1} ~\right]^2 \implies $$

$$a_{n+1} - 2a_n = \sqrt{3a_n^2 - 2} = 2a_n - a_{n-1}.$$

Note
I started by trying to prove the assertion by induction. I then accidentally backed in to a direct algrebraic proof that doesn't rely on induction.

That is, I (inadvertantly) made no use of the inductive hypothesis that (for example) $~a_n = 4a_{n-1} - a_{n-2}.~$ Instead, I simply proved the assertion directly from the $~\displaystyle a_{n+1} - 2a_n = \sqrt{3a_n^2 - 2}~$ specification.

Minor Point
Directly from the specification, you have that every element in the $~\langle a_n\rangle~$ sequence is positive.

Also, directly from the specification, you have that the sequence is strictly increasing.

These two points justify the assumption made in my analysis that $~2a_n - a_{n-1} > 0.~$