Question

Write the sum using summation notation and then find the sum: 12, -4, 4/3, -4/9, ...

Answer 1

So I've got \[\sum_{n = 1}^{\infty} 12(\frac{ -1 }{ 3 })^{n-1}\]

and the equation for a sum of an infinite series is A\[S _{\infty} = \frac{ A _{1} }{ r - 1 }\] which I have to equal 9. Is that right?

Answer 2

\[S _{\infty}=\frac{ a }{ 1-r }=\frac{ 12 }{ 1-(\frac{ -1 }{ 3 } )}=12 \times \frac{ 3 }{ 4 }=9\]

Answer 3

@sshayer So I am correct?

Answer 4

Yes, you are to find the sum of an infinite geometric series. Thanks for sharing your work. Looks good!

Answer 5

remember,in the denominator 1-r

Answer 6

You've done fine. But for communication's sake, identify your "first term:" a = 12, and your "common ratio:" r =-1/3.

Answer 7

Right

Answer 8

Yeah I meant to say 1-r.