Step 1. The sequence that we have in the table is:
![5,\text{ 10, 20, ...}](https://img.qammunity.org/2023/formulas/mathematics/college/k3eqnpvhie9s3ijhys938om13235pw1kqj.png)
As you can see the number doubles each time.
We require to find the expression that represents the sum for term 3 through term 9 in sigma notation.
Step 2. First, since the summation has to be from term 3 to term 9, the sigma notation should look as follows:
![\sum_{n\mathop{=}3}^9](https://img.qammunity.org/2023/formulas/mathematics/college/uoqdfg9si53y7h19w9onoajce6od7k76bs.png)
This discards options 1 and 3.
Step 3. Now we need to find an expression that represents the sum of the terms. If we continue the sequence the numbers would be:
![5,10,20,40,80,...](https://img.qammunity.org/2023/formulas/mathematics/college/85hw3tfcj8pvazjgs0b0qu89izwwj6c2dp.png)
We can also express this as 5 multiplied by a power of 2:
![5\cdot2^0+5\cdot2^1+5\cdot2^3+...](https://img.qammunity.org/2023/formulas/mathematics/college/7kwv6bsvaaqvsrz0j2n0exfywhnh4y7a3g.png)
That is because
2^0=1
2^1=2
2^2^4
.
.
.
Therefore, the result of the multiplications:
![undefined]()
This can be simplified to:
![5(2)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/suzb7fx34ieje4nmwwn2v44iwvmq6r5s8f.png)
Step 4. The final expression is:
![\sum_{n\mathop{=}3}^95(2)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/jzmqcj1uulfav25rhg01qmrtt9kutlbo6t.png)
Which is shown in the second option.
Answer:
![\sum_{n\mathop{=}3}^95(2)^(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/jzmqcj1uulfav25rhg01qmrtt9kutlbo6t.png)