Given 𝑆𝑛 = 3𝑛+2 − 9. Show that 𝑇𝑛 is given by 𝑇𝑛 = 2(3 𝑛+1)

To show that \( T_n \) is given by \( T_n = 2(3^{n+1}) \) given \( S_n = 3^n + 2 - 9 \), we can proceed as follows:

### 1. Understand the relationship between \( S_n \) and \( T_n \)
Assume \( T_n \) represents the \(n\)-th term of a sequence. The sum of the first \( n \) terms of this sequence is denoted by \( S_n \), so we have:
\[
S_n = T_1 + T_2 + \cdots + T_n
\]

The relationship between consecutive sums is given by:
\[
T_n = S_n - S_{n-1}
\]

### 2. Find \( S_{n-1} \)
Given \( S_n = 3^n + 2 - 9 \), we first need to express \( S_{n-1} \):
\[
S_{n-1} = 3^{n-1} + 2 - 9
\]

### 3. Find the expression for \( T_n \)
Using the relation \( T_n = S_n - S_{n-1} \), substitute the expressions for \( S_n \) and \( S_{n-1} \):
\[
T_n = \left(3^n + 2 - 9\right) - \left(3^{n-1} + 2 - 9\right)
\]

Simplify the equation:
\[
T_n = 3^n - 3^{n-1}
\]

### 4. Factor the expression
Factor out \( 3^{n-1} \) from the equation:
\[
T_n = 3^{n-1}(3 - 1)
\]

This simplifies to:
\[
T_n = 2 \times 3^{n-1} \times 3 = 2 \times 3^{n+1}
\]

Therefore, the expression for \( T_n \) is:
\[
T_n = 2 \times 3^{n+1}
\]

### Conclusion
We have shown that \( T_n = 2(3^{n+1}) \) as required.

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Do you want more details or have any questions about this? Here are some related questions:

1. How do we derive the general term of a sequence from its sum?
2. What is the meaning of the term "consecutive sum" in sequences?
3. Can \( S_n \) and \( T_n \) be used to determine the sum of the first \( n \) terms in any sequence?
4. How would this process change if the formula for \( S_n \) was different?
5. How can we verify the solution using specific values of \( n \)?

**Tip:** When dealing with sequences, always check the relation between terms and sums carefully to avoid errors in subtraction or factoring.

Learn how to prove Tn = 2(3^(n+1)) using the relationship between Sn = 3^n + 2 - 9 and Tn. This problem involves sequences, summation formulas, and exponential functions. Suitable for high school students.

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