You provided a sigma notation problem involving a geometric sequence. The sum is described as having 6 terms, with $N = 1$ as the lower limit, and the formula is $7^{3-i}$.

To express this properly:

$\sum_{i=1}^{6} 7^{3-i}$

This summation involves computing the sum of 6 terms, where each term is $7^{3-i}$ for values of $i$ from 1 to 6.

Steps to solve:

1. Evaluate each term of the sequence:

   • For $i = 1$: $7^{3-1} = 7^2 = 49$
   • For $i = 2$: $7^{3-2} = 7^1 = 7$
   • For $i = 3$: $7^{3-3} = 7^0 = 1$
   • For $i = 4$: $7^{3-4} = 7^{-1} = \frac{1}{7}$
   • For $i = 5$: $7^{3-5} = 7^{-2} = \frac{1}{49}$
   • For $i = 6$: $7^{3-6} = 7^{-3} = \frac{1}{343}$

2. Sum these values:

$49 + 7 + 1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343}$

Now let's compute the sum numerically:

$\sum_{i=1}^{6} 7^{3-i} = 49 + 7 + 1 + 0.14286 + 0.02041 + 0.00292$

$= 57.16619$

Thus, the sum of the series is approximately $57.17$.

Would you like further details or a breakdown of each step?

5 Relative Questions:

1. How is sigma notation applied in arithmetic or geometric sequences?
2. What is the difference between arithmetic and geometric sequences?
3. How do negative exponents affect the behavior of a geometric sequence?
4. Can sigma notation be used for more complex series, like alternating or exponential sums?
5. How would you handle infinite geometric series in sigma notation?

Tip:

When summing terms with negative exponents, it's helpful to express them as fractions for clearer understanding and accurate calculation.