Question

Part 1: Determine the geometric partial sum (see below) .

Part 2: In complete sentences, explain the necessary steps required to find the partial sum. Include an explanation of whether this partial sum is divergent or convergent in nature.

Answer 1

\[\sum_{i=1}^{7} 3(2)^{i-1}\]

Answer 2

First part, only the value needed?
then: "your sum" = \(\frac32\sum_{i=1}^72^i = \frac32 (2+4+8+16+32+64+128) = ...\)

Do you need to use the formula \(\sum_{k=0}^n x^k = \frac{1-x^{n+1}}{1-x}\) instead?

Answer 3

I'm not sure, think so

Answer 4

you manipulate the expression \(\sum_{i=1}^7 32^{i-1}\) and obtain \(3 \sum_{i=1}^72^{i-1}\). Here you can work on the summing index. if i=1,2,...,7, but you always use the value i-1, then you can choose the summing index j=i-1, which starts at 0 and ends at 6. Starting at 0 is good because you can apply the formula.

Answer 5

I don't know

Answer 6

dont know what?

Answer 7

how to get the answer

Answer 8

\(3\sum_{i=1}^72^{i-1} = 3\sum_{j=0}^6 2^j\) (where j=i-1)
\(= 3 \frac{1-2^{6+1}}{1-2}\) (using the formula)
\( = 3(2^{6+1}-1) = 3\times 6^7 - 3\).

replace 7 by \(n\). the answer would be \(3\times 6^n - 3\). This is the value of the partial sum. Then it seems this quantity always increases as \(n\to\infty\).

Answer 9

okay so back space this. the first thing that you wrote to me makes more sense with what I'm studying