Sq. Sequences - JulTob/Mathematics GitHub Wiki

The sequence $( {a_n} )$ is defined inductively with the following properties:

  1. Base Cases:
   a_1 = 1 \quad \text{and} \quad a_2 = 1.
  1. Recursive Relation:
   a_{n+2} = 2a_{n+1} - a_n \quad \text{for } n \geq 1.

Let's analyze this sequence further to understand its behavior and find additional terms. We’ll compute the first few terms of the sequence to identify any potential pattern.

Calculating the First Few Terms

Using the recursive relation $( a_{n+2} = 2a_{n+1} - a_n )$:

  1. We already know:
   a_1 = 1, \quad a_2 = 1.
  1. Calculate $( a_3 )$:
   a_3 = 2a_2 - a_1 = 2 \cdot 1 - 1 = 2 - 1 = 1.
  1. Calculate $( a_4 )$:
   a_4 = 2a_3 - a_2 = 2 \cdot 1 - 1 = 2 - 1 = 1.
  1. Calculate $( a_5 )$:
   a_5 = 2a_4 - a_3 = 2 \cdot 1 - 1 = 2 - 1 = 1.
  1. Calculate $( a_6 )$:
   a_6 = 2a_5 - a_4 = 2 \cdot 1 - 1 = 2 - 1 = 1.

Observing the Pattern

From these calculations, we can observe that every term calculated so far is equal to $( 1 )$. This suggests that the sequence might be constant with $( a_n = 1 )$ for all $( n \geq 1 )$.

Proving the Pattern by Induction

To verify that $( a_n = 1 )$ for all $( n \geq 1 )$, we can use mathematical induction.

Step 1: Base Cases

We know that:

a_1 = 1 \quad \text{and} \quad a_2 = 1,

which matches our claim.

Step 2: Inductive Hypothesis

Assume that $( a_k = 1 )$ and $( a_{k+1} = 1 )$ for some $( k \geq 1 )$.

Step 3: Inductive Step

We need to show that $( a_{k+2} = 1 )$ under this assumption. Using the recursive relation:

a_{k+2} = 2a_{k+1} - a_k.

Substituting $( a_{k+1} = 1 )$ and $( a_k = 1 )$ (from our inductive hypothesis), we get:

a_{k+2} = 2 \cdot 1 - 1 = 2 - 1 = 1.

Thus, if $( a_k = 1 )$ and $( a_{k+1} = 1 )$, then $( a_{k+2} = 1 )$.

Conclusion

By the principle of mathematical induction, we have shown that $( a_n = 1 )$ for all $( n \geq 1 )$.

Final Answer

\boxed{a_n = 1 \text{ for all } n \geq 1}.

To prove that $( a_n = 3 + 2^n )$ for all natural numbers $( n )$ in the sequence defined by:

  1. Base Cases:
   a_1 = 5, \quad a_2 = 7.
  1. Recursive Relation:
   a_{n+2} = 3a_{n+1} - 2a_n.

We’ll use mathematical induction to prove that $( a_n = 3 + 2^n )$ for all $( n \geq 1 )$.

Step 1: Base Cases

First, let’s verify that the formula $( a_n = 3 + 2^n )$ holds for $( n = 1 )$ and $( n = 2 )$.

  1. For $( n = 1 )$:
   a_1 = 5.

The formula gives $( a_1 = 3 + 2^1 = 3 + 2 = 5 )$, which matches $( a_1 = 5 )$.

  1. For $( n = 2 )$:
   a_2 = 7.

The formula gives $( a_2 = 3 + 2^2 = 3 + 4 = 7 )$, which matches $( a_2 = 7 )$.

Thus, the formula holds for the base cases $( n = 1 )$ and $( n = 2 )$.

Step 2: Inductive Hypothesis

Assume that the formula holds for some positive integer $( k )$; that is, assume:

a_k = 3 + 2^k \quad \text{and} \quad a_{k+1} = 3 + 2^{k+1}.

This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the formula holds for $( a_k )$ and $( a_{k+1} )$, then it also holds for $( a_{k+2} )$. In other words, we need to show:

a_{k+2} = 3 + 2^{k+2}.

Proof of the Inductive Step

Using the recursive relation, we have:

a_{k+2} = 3a_{k+1} - 2a_k.

Substitute the expressions for $( a_k )$ and $( a_{k+1} )$ from the inductive hypothesis:

a_{k+2} = 3(3 + 2^{k+1}) - 2(3 + 2^k).

Now expand and simplify:

a_{k+2} = 3 \cdot 3 + 3 \cdot 2^{k+1} - 2 \cdot 3 - 2 \cdot 2^k.

a_{k+2} = 9 + 3 \cdot 2^{k+1} - 6 - 2 \cdot 2^k.

a_{k+2} = 3 + 3 \cdot 2^{k+1} - 2 \cdot 2^k.

Now, rewrite $( 3 \cdot 2^{k+1} )$ as $( 2^{k+2} + 2^{k+1} )$:

a_{k+2} = 3 + 2^{k+2} + 2^{k+1} - 2 \cdot 2^k.

Notice that $( 2^{k+1} - 2 \cdot 2^k = 0 )$, so we are left with:

a_{k+2} = 3 + 2^{k+2}.

This matches the formula we set out to prove for $( a_{k+2} )$.

Conclusion

Since we have shown that:

  1. The formula $( a_n = 3 + 2^n )$ holds for the base cases $( n = 1 )$ and $( n = 2 )$,
  2. If the formula holds for $( n = k )$ and $( n = k + 1 )$, then it also holds for $( n = k + 2 )$,

we conclude by the principle of mathematical induction that the formula is true for all natural numbers $( n )$:

\boxed{a_n = 3 + 2^n \text{ for all } n \geq 1}.