Question

Stuck on this problem! Help please!
Will give medal.

Answer 1

@jdoe0001 Please help!

Answer 2

\(\Large S_{\color{brown}{ n}}\implies {\huge \Sigma}_{i=0}^{\color{brown}{ n}}\quad a_1\cdot r^{{\color{brown}{ n}}-1}\implies a_1\left(\cfrac{1-r^{\color{brown}{ n}}}{1-r}\right) \)

can you see what that will be?

Answer 3

hmm I kinda threw in "i" there =) \(\Large S_{\color{brown}{ n}}\implies {\huge \Sigma}_{i=0}^{\color{brown}{ n}}\quad a_1\cdot r^{i-1}\implies a_1\left(\cfrac{1-r^{\color{brown}{ n}}}{1-r}\right)\)

Answer 4

hm, I see you are making a infinite sum. I don't understand how to answer the second question though :(

Answer 5

hmmm wel.... actually "i" should start off 1 anyhow \(\Large S_{\color{brown}{ n}}\implies {\huge \Sigma}_{i=1}^{\color{brown}{ n}}\quad a_1\cdot r^{i-1}\implies a_1\left(\cfrac{1-r^{\color{brown}{ n}}}{1-r}\right)\)

Answer 6

so... we want to know the finite sum up to the 20th term.... and we know the 1st term, 3
if you look at the expression.... our "r" is 1/2 so

Answer 7

\(\Large S_{\color{brown}{ 20}}\implies {\huge \Sigma}_{i=1}^{\color{brown}{ 20}}\quad a_1\cdot r^{i-1}\implies 3\left(\cfrac{1-\frac{1}{2}^{\color{brown}{ 20}}}{1-\frac{1}{2}}\right)\)

Answer 8

Okay so If I solve that out that would find the answer right? :)

Answer 9

well... I guess I should enclose the 1/2 in the exponent so \(\Large S_{\color{brown}{ 20}}\implies {\huge \Sigma}_{i=1}^{\color{brown}{ 20}}\quad a_1\cdot r^{i-1}\implies 3\left(\cfrac{1-\left(\frac{1}{2}\right)^{\color{brown}{ 20}}}{1-\frac{1}{2}}\right)\)

Answer 10

yeap that'd give the "finite sum" up to the 20th term

Answer 11

I got 2.499997139. I don't know if that is completely right though.

Answer 12

I just did the the calculations again and got 5.99999. @jdoe0001

Answer 13

yeap I got 5.9999942779541015625

Answer 14

@jdoe0001 Thank you very much. Sorry I didn't close the parenthesis first.

Answer 15

yw