Use a graphing calculator to find the following sum. \( \sum_{n=1}^{7} 3^{n-2} 3^{n-2}=\square \) (Type an integer or decimal rounded to three decimal places as needed.
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To solve the sum \( \sum_{n=1}^{7} 3^{n-2} 3^{n-2} \), we can simplify the expression first. The term \( 3^{n-2} 3^{n-2} \) can be rewritten as \( (3^{n-2})^2 = 3^{2(n-2)} = 3^{2n-4} \). Thus, our sum becomes: \[ \sum_{n=1}^{7} 3^{2n-4} \] This can also be expressed as: \[ \sum_{n=1}^{7} 3^{2n-4} = \sum_{n=1}^{7} \frac{3^{2n}}{3^4} = \frac{1}{81} \sum_{n=1}^{7} 3^{2n} \] Next, we compute \( \sum_{n=1}^{7} 3^{2n} \). This is a geometric series where the first term \( a = 3^2 = 9 \) and the common ratio \( r = 3^2 = 9 \). For a geometric series, the sum can be calculated using the formula: \[ S_n = a \frac{1-r^n}{1-r} \] Applying this to our series: \[ S_7 = 9 \frac{1-9^7}{1-9} = 9 \frac{1-4782969}{-8} = 9 \cdot \frac{-4782968}{-8} = 9 \cdot 597871 = 5389839 \] Now we plug this back into our earlier equation: \[ \sum_{n=1}^{7} 3^{2n-4} = \frac{1}{81} \cdot 5389839 \] Thus, \[ \sum_{n=1}^{7} 3^{2n-4} = 66562.543 \] Therefore, the result of the sum \( \sum_{n=1}^{7} 3^{n-2} 3^{n-2} \) is approximately: \[ \boxed{66562.543} \]
