To solve the given summation:

$\sum_{k=5}^{11} 3^k$

Step 1: Identify the Series

The summation involves the powers of $3$, starting at $k = 5$ and ending at $k = 11$. This is a geometric series.

The terms are: $3^5, 3^6, 3^7, 3^8, 3^9, 3^{10}, 3^{11}.$

Step 2: General Formula for a Finite Geometric Series

The sum of a geometric series is given by the formula:

$S = a \frac{r^n - 1}{r - 1},$ where:

• $a$ is the first term,
• $r$ is the common ratio,
• $n$ is the number of terms.

Here:

• $a = 3^5$ (the first term),
• $r = 3$ (the common ratio),
• $n = 11 - 5 + 1 = 7$ (number of terms).

Step 3: Substitute Values into the Formula

The first term is $a = 3^5 = 243$, the ratio is $r = 3$, and the number of terms is $n = 7$.

$S = 243 \cdot \frac{3^7 - 1}{3 - 1}.$

Simplify step-by-step:

• $3^7 = 2187$,
• The denominator $3 - 1 = 2$.

Substitute back: $S = 243 \cdot \frac{2187 - 1}{2}.$

Simplify further: $S = 243 \cdot \frac{2186}{2}.$ $S = 243 \cdot 1093.$

Step 4: Multiply

Now calculate $243 \times 1093$:

$243 \times 1093 = 265599.$

Final Answer:

$\sum_{k=5}^{11} 3^k = 265599.$

Do you need more explanation or help with any part of this solution?

Related Questions:

1. How do you solve a geometric series where the ratio $r$ is less than 1?
2. What is the formula for the sum of an infinite geometric series?
3. How do you calculate the sum of an arithmetic series?
4. How do you simplify powers of a number using exponent rules?
5. Can the geometric series formula be derived?

Tip: When working with geometric series, always check that the ratio $r \neq 1$ to avoid undefined cases.