1,5,13,29,61,125,... find the recursive formulae

The given sequence is:  
1, 5, 13, 29, 61, 125, ...

To find the recursive formula, we need to observe the relationship between consecutive terms.

Let's denote the terms of the sequence as \( a_n \), where \( a_1 = 1 \), \( a_2 = 5 \), \( a_3 = 13 \), and so on.

### Step 1: Calculate the differences between consecutive terms:
\[
a_2 - a_1 = 5 - 1 = 4
\]
\[
a_3 - a_2 = 13 - 5 = 8
\]
\[
a_4 - a_3 = 29 - 13 = 16
\]
\[
a_5 - a_4 = 61 - 29 = 32
\]
\[
a_6 - a_5 = 125 - 61 = 64
\]

This shows that the differences between consecutive terms are: 4, 8, 16, 32, 64,...

### Step 2: Observe the pattern in the differences:
The differences form a geometric sequence where each term is twice the previous one. That is:
\[
\Delta a_n = 2^{n+1}
\]

### Step 3: Establish the recursive relation:
Since the difference between consecutive terms follows a pattern, the recursive formula for the sequence is:

\[
a_{n+1} = a_n + 2^{n+1}
\]
with the initial condition:
\[
a_1 = 1
\]

This is the recursive formula for the given sequence.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How would you derive the explicit formula for this sequence?
2. What is the sum of the first \( n \) terms of this sequence?
3. How can the behavior of the geometric difference be generalized for other sequences?
4. Can you find a different recursive relation for other similar sequences?
5. How does this recursive relation help in finding terms for large \( n \)?

**Tip:** Recursive formulas are often easier to derive and work with when there's a clear pattern in differences between consecutive terms.

Discover how to find the recursive formula for the sequence 1, 5, 13, 29, 61, 125,... by identifying the pattern in the differences between terms. Learn about geometric progressions and recursive relations in this solution. Suitable for students in Grades 9-12.

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