Step 1. The sequence that we have in the table is:
[tex]5,\text{ 10, 20, ...}[/tex]
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:
[tex]\sum_{n\mathop{=}3}^9[/tex]
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:
[tex]5,10,20,40,80,...[/tex]
We can also express this as 5 multiplied by a power of 2:
[tex]5\cdot2^0+5\cdot2^1+5\cdot2^3+...[/tex]
That is because
2^0=1
2^1=2
2^2^4
.
.
.
Therefore, the result of the multiplications:
[tex]undefined[/tex]
This can be simplified to:
[tex]5(2)^{n-1}[/tex]
Step 4. The final expression is:
[tex]\sum_{n\mathop{=}3}^95(2)^{n-1}[/tex]
Which is shown in the second option.
Answer:
[tex]\sum_{n\mathop{=}3}^95(2)^{n-1}[/tex]
