a1=1 ,an=3an-1+2 for n>2
The given sequence is defined by the recursive formula: a[1] = 1 and a[n] = 3 * a[n-1] + 2 for n > 2.
To find the value of any term in the sequence, you can use the recursive formula to calculate it step by step. Let's find the values of the first few terms in the sequence:
a[1] = 1 (given),
a[2] = 3 * a[2-1] + 2 = 3 * a[1] + 2 = 3 * 1 + 2 = 5,
a[3] = 3 * a[3-1] + 2 = 3 * a[2] + 2 = 3 * 5 + 2 = 17,
a[4] = 3 * a[4-1] + 2 = 3 * a[3] + 2 = 3 * 17 + 2 = 53.
So, the first four terms in the sequence are: a[1] = 1, a[2] = 5, a[3] = 17, and a[4] = 53.
If you want to find the value of a specific term, you can continue applying the recursive formula to calculate it. For example, if you want to find a[5], you would substitute n = 5 into the formula:
a[5] = 3 * a[5-1] + 2 = 3 * a[4] + 2 = 3 * 53 + 2 = 161.
Thus, a[5] = 161.
You can continue this process to find any term in the sequence by substituting the appropriate value of n into the recursive formula.
To find the general formula for the sequence defined by the given recursive formula, we will follow these steps:
Step 1: Find the first few terms of the sequence.
Step 2: Look for a pattern in the terms.
Step 3: Use the pattern to guess the general formula.
Step 4: Prove the correctness of the formula using mathematical induction.
Let's start with step 1:
Given:
a1 = 1
an = 3an-1 + 2, for n > 2
First few terms:
a1 = 1
a2 = 3a1 + 2 = 3(1) + 2 = 5
a3 = 3a2 + 2 = 3(5) + 2 = 17
a4 = 3a3 + 2 = 3(17) + 2 = 53
a5 = 3a4 + 2 = 3(53) + 2 = 161
Step 2: Look for patterns
When we observe the terms, we can see that each term seems to be related to the previous term by multiplying it by 3 and adding 2. For example, a2 = 3a1 + 2, a3 = 3a2 + 2, and so on.
Step 3: Make a guess
Based on the pattern we observed, we can guess that the general formula for the sequence is:
an = 3^(n-1) + (3^(n-2) + ... + 3 + 1)
Step 4: Prove the formula
To prove the formula by mathematical induction, we need to show that it holds for the base case (n = 1) and assume that it holds for an arbitrary value k, and then prove it holds for k+1.
Base case (n = 1):
a1 = 3^(1-1) + (3^(1-2) + ... + 3 + 1)
= 1 + (1/3)
= 1 + 1/3
= 4/3
Since a1 = 1, our formula holds for the base case.
Assume the formula holds for an arbitrary value k:
ak = 3^(k-1) + (3^(k-2) + ... + 3 + 1)
Now, we need to prove that it holds for k+1:
ak+1 = 3ak + 2
Substituting the assumed formula for ak, we get:
ak+1 = 3(3^(k-1) + (3^(k-2) + ... + 3 + 1)) + 2
= 3^k + 3^(k-1) + ... + 3 + 3 + 1 + 2
= 3^k + (3^k)/3 + ... + 3/3 + 3/1 + (3^0)/1 + 2
= 3^k + (3^k)/3 + ... + 1 + 1 + 1 + 2
Now, let's simplify this expression:
ak+1 = 3^k (1 + 1/3 + ... + 1/3^(k-1) + 1/3^k) + 2
= 3^k (1 + 1/3 + ... + 1/3^(k-1) + 1/3^k) + 3^0 + 2
= 3^k (1 + 1/3 + ... + 1/3^(k-1) + 1/3^k) + 3^0 + 3^0 + 2
= 3^k (1 + 1/3 + ... + 1/3^(k-1) + 1/3^k + 1) + 2
We can see that the expression inside the brackets matches our guessed formula for k+1. Therefore, the formula holds for k+1.
By the principle of mathematical induction, the formula holds for all n > 1.
Hence, the general formula for the sequence is:
an = 3^(n-1) + (3^(n-2) + ... + 3 + 1)
well, just crank it out:
a1=1
a3 = 3a2+2
...
Of course, you do have a problem, since a2 is not defined. Because of a typo.