Geometric Determine the value of P if $$ \mathrm{P}=\sum_{k=1}^{13} 3^{k-5} $$ rove that for any arithmetic sequence of which the first term is $$ a $$ and the constant lifference is $$ d $$, the sum to $$ n $$ terms can be expressed as $$ S_{n}=\frac{n}{2}(2 a+(n-1) d) $$.

Question

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Certainly! Let's address both parts of your query step by step.

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## **1. Determining the Value of $$ \mathrm{P} = \sum_{k=1}^{13} 3^{k-5} $$**

### **Understanding the Series**

The given sum is a geometric series where each term is of the form $$ 3^{k-5} $$ for $$ k $$ ranging from 1 to 13.

### **Rewriting the Series**

First, let's simplify the expression inside the summation:

$$
3^{k-5} = \frac{3^k}{3^5} = \frac{1}{3^5} \cdot 3^k
$$

Therefore, the sum $$ P $$ can be rewritten as:

$$
\mathrm{P} = \sum_{k=1}^{13} 3^{k-5} = \frac{1}{3^5} \sum_{k=1}^{13} 3^k
$$

### **Calculating the Geometric Series**

The sum $$ \sum_{k=1}^{13} 3^k $$ is a geometric series with:
- **First term ($$ a $$)**: $$ 3^1 = 3 $$
- **Common ratio ($$ r $$)**: 3
- **Number of terms ($$ n $$)**: 13

The sum of the first $$ n $$ terms of a geometric series is given by:

$$
S_n = a \cdot \frac{r^n - 1}{r - 1}
$$

Plugging in the values:

$$
\sum_{k=1}^{13} 3^k = 3 \cdot \frac{3^{13} - 1}{3 - 1} = 3 \cdot \frac{3^{13} - 1}{2}
$$

### **Final Calculation of $$ P $$**

Now, substitute back to find $$ P $$:

$$
\mathrm{P} = \frac{1}{3^5} \cdot 3 \cdot \frac{3^{13} - 1}{2} = \frac{3^{13} - 1}{2 \cdot 3^4}
$$

Simplify $$ 3^{13} / 3^4 $$:

$$
3^{13} / 3^4 = 3^{9} = 19,683
$$

Thus,

$$
\mathrm{P} = \frac{19,683 - 1}{2} = \frac{19,682}{2} = 9,841
$$

**Answer:** $$ \mathrm{P} = 9,\!841 $$

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## **2. Proving the Sum Formula for an Arithmetic Sequence**

### **Given:**

An arithmetic sequence with:
- **First term ($$ a $$)**
- **Common difference ($$ d $$)**
- **Number of terms ($$ n $$)**

### **To Prove:**

$$
S_n = \frac{n}{2} \left( 2a + (n - 1)d \right)
$$

### **Proof:**

Consider the arithmetic sequence:

$$
a, \quad a + d, \quad a + 2d, \quad \ldots, \quad a + (n-1)d
$$

**Sum of the first $$ n $$ terms ($$ S_n $$):**

$$
S_n = a + (a + d) + (a + 2d) + \ldots + \left(a + (n - 1)d\right)
$$

**Writing the sum in reverse order:**

$$
S_n = \left(a + (n - 1)d\right) + \left(a + (n - 2)d\right) + \ldots + (a + d) + a
$$

**Adding the two expressions term by term:**

$$
2S_n = [a + (a + (n - 1)d)] + [(a + d) + (a + (n - 2)d)] + \ldots + [a + (a + (n - 1)d)]
$$

Each pair sums to:

$$
a + a + (n - 1)d = 2a + (n - 1)d
$$

Since there are $$ n $$ such pairs (because each term is paired once):

$$
2S_n = n \left( 2a + (n - 1)d \right)
$$

**Solving for $$ S_n $$:**

$$
S_n = \frac{n}{2} \left( 2a + (n - 1)d \right)
$$

**Thus, the formula is proven.**

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**Conclusion:** We've determined that $$ \mathrm{P} = 9,\!841 $$ and successfully proved the sum formula for an arithmetic sequence.
$$ \mathrm{P} = 9,\!841 $$ For any arithmetic sequence with first term $$ a $$ and common difference $$ d $$, the sum of the first $$ n $$ terms is $$ S_n = \frac{n}{2} (2a + (n - 1)d) $$.

Geometric
Determine the value of P if
rove that for any arithmetic sequence of which the first term is and the constant
lifference is , the sum to terms can be expressed as .

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Solución

1. Determining the Value of

Understanding the Series

Rewriting the Series

Calculating the Geometric Series

Final Calculation of

2. Proving the Sum Formula for an Arithmetic Sequence

Given:

To Prove:

Proof:

Beyond the Answer

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Geometric Determine the value of P if rove that for any arithmetic sequence of which the first term is and the constant lifference is , the sum to terms can be expressed as .