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Mathematics 14 Online
OpenStudy (anonymous):

What is the sum of the geometric series ? I attached an image. Its in sigma notation

OpenStudy (anonymous):

OpenStudy (rational):

I think you need to use "geometric series partial sum formula"

OpenStudy (rational):

look it up in your notes and see if you can find it..

OpenStudy (anonymous):

okay!

OpenStudy (anonymous):

An=A1+(n-1) -2

OpenStudy (rational):

nope

OpenStudy (anonymous):

ohh wait thats for arithmetic right?

OpenStudy (rational):

exactly thats for airthmetic sequence but we want "geometric series partial sum" formula

OpenStudy (anonymous):

sn= a1*(1-r^n)/(1-r)

OpenStudy (rational):

Yes thats the one

OpenStudy (rational):

\[S_n = a_1 \dfrac{1-r^n}{1-r}\] find out the values and plug in

OpenStudy (anonymous):

im not sure what i would plug in for r

OpenStudy (rational):

we can eyeball by comparing with : \[\sum\limits_{n=0}^M a(r)^n\] we get \(r = 3/4\)

OpenStudy (rational):

and you get first term by plugging in n=0 : \[3\left(\dfrac{3}{4}\right)^0 = 3(1) = 3\] so \(a =3 \)

OpenStudy (rational):

since there are 19 terms, we have \(n=19\)

OpenStudy (rational):

then the partial sum would be \[\large 3\dfrac{1- \left(\frac{3}{4}\right)^{19}}{1-\frac{3}{4}}\] simplify

OpenStudy (anonymous):

11.94 ?

OpenStudy (anonymous):

Thank you so much this makes so much more sense now !

OpenStudy (campbell_st):

umm there is an error in the solution, the values of n finish at 18... and not 19 as shown in the formula

OpenStudy (anonymous):

thanks but to find n( the numbers in the solution) , you subtract the limits & add one

OpenStudy (anonymous):

so n= 19

OpenStudy (anonymous):

@campbell_st

OpenStudy (rational):

Yep! im getting 11.949 which rounds to 11.9\(\color{Red}{5}\)

OpenStudy (anonymous):

awesome ! thanks again!

OpenStudy (rational):

yw!

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