If a, start subscript, 1, end subscript, equals, 3a 
1
​
 =3 and a, start subscript, n, end subscript, equals, left parenthesis, a, start subscript, n, minus, 1, end subscript, right parenthesis, squared, plus, na 
n
​
 =(a 
n−1
​
 ) 
2
 +n then find the value of a, start subscript, 3, end subscripta 
3
​
 .

To solve for \( a_3 \), we need to find the terms of the sequence using the given formula:

1. We are given:
   \[
   a_1 = 3
   \]

2. The recursive formula is:
   \[
   a_n = (a_{n-1})^2 + n
   \]

We need to find \( a_3 \). 

### Step 1: Find \( a_2 \)

Using the formula for \( n = 2 \):

\[
a_2 = (a_1)^2 + 2
\]

Substitute \( a_1 = 3 \):

\[
a_2 = (3)^2 + 2 = 9 + 2 = 11
\]

### Step 2: Find \( a_3 \)

Now, using the formula for \( n = 3 \):

\[
a_3 = (a_2)^2 + 3
\]

Substitute \( a_2 = 11 \):

\[
a_3 = (11)^2 + 3 = 121 + 3 = 124
\]

### Conclusion

The value of \( a_3 \) is:
\[
\boxed{124}
\]

Would you like more details on the steps, or have any questions?

### Related Questions:

1. What is the general formula for finding \( a_n \) in this sequence?
2. How would the sequence change if the initial value \( a_1 \) was different?
3. Can you find \( a_4 \) using the same recursive method?
4. What are some common properties of sequences defined by recursive formulas?
5. How does this sequence behave as \( n \) becomes very large?

**Tip:** In recursive sequences, always start with the initial value and proceed step-by-step to find the next terms.

If a₁ = 3 and aₙ = (aₙ₋₁)² + n, then find the value of a₃.

Learn how to find a₃ in a recursive sequence with a₁ = 3 and aₙ = (aₙ₋₁)² + n. This detailed solution covers recursive sequences, exponents, and algebraic concepts, suitable for students in Grades 9-11.

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