Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

Find the sum of the first 20 terms of the geometric series -2 + 4 + -8 + ....

OpenStudy (anonymous):

The first term of the GP is -2 and the ratio is -2. What is the formula that links all of this.

OpenStudy (anonymous):

lol idk

OpenStudy (anonymous):

please dont comment if you arent gonna help. @callme911

OpenStudy (australopithecus):

sorry that would be \[\sum_{n=1}^{20}(-1)^n 2^n\]

OpenStudy (australopithecus):

do you understand how I came to this solution?

OpenStudy (australopithecus):

or what to do next?

OpenStudy (anonymous):

whyd you delete it?

OpenStudy (anonymous):

Lol....u see it is G.P with r=-2 \[an=a*r ^{n-1}\]

OpenStudy (anonymous):

\[-2*-2^{19}\]

OpenStudy (anonymous):

\[=-2^{20}\]

OpenStudy (anonymous):

Hope that Helps

OpenStudy (australopithecus):

why did I delete what?

OpenStudy (anonymous):

no.. i do not

OpenStudy (australopithecus):

whenever you have a problem that alternates from negative to positive you will have something like (-1)^n if it starts out as a positive then you will have (-1)^(n+1)

OpenStudy (australopithecus):

remember in the case of (-1)^n when n is even, (-1)^n equals a positive when n is odd, (-1)^n equals a negative

OpenStudy (australopithecus):

so from the series I gave you \[\sum_{n}^{\infty} (-1)^n (2)^n\] = (-1)^(1)(2^1) + (-1)^2(2^2) + (-1)^(3)(2^3) + (-1)^(4)(2^4)... -2 + 4 + -8 + 16...

OpenStudy (australopithecus):

do that 20 times and add them all and you have your answer

OpenStudy (australopithecus):

I think there may be a formula for these questions that makes it easier but I do not know it

OpenStudy (anonymous):

Lol.....@Australopithecus no need to Do...that....we have Formulas

OpenStudy (australopithecus):

yeah I know yahoo!

OpenStudy (australopithecus):

I just dont know it lol

OpenStudy (australopithecus):

plus I wanted to demonstrate the operation

OpenStudy (australopithecus):

also at the bottom of that summation it should be n=1 not just n

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!