infinite series problem

**dylan5188** · April 1st 2010, 05:41 PM

Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation. A.

I got -4 which is wrong.
Any thought on that one ?

**skeeter** · April 1st 2010, 05:45 PM

Originally Posted by dylan5188

Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation. A.

I got -4 which is wrong.
Any thought on that one ?

you found the sum of an infinite geometric series ... this one is finite.

**dylan5188** · April 1st 2010, 05:51 PM

Originally Posted by skeeter

you found the sum of an infinite geometric series ... this one is finite.

yea I used way from textbook. In this problem, r is -1/2. which has aabs value<1.
and put in to a/(1-r) then I got -4.

**Anonymous1** · April 1st 2010, 05:53 PM

$-6+3-3(\frac{1}{2}+...+\frac{1}{2^{11}} ) = -3(1+(\frac{1}{2} +...+\frac{1}{2^{11}} ))= -3(1+(2^{10} +...+2+1)/2^{11})...$

Figure out what $2^{10} + 2^9+...+2 + 1$ is and your pretty much done.

**dylan5188** · April 1st 2010, 06:10 PM

Originally Posted by Anonymous1

$-6+3-3(\frac{1}{2}+...+\frac{1}{2^{11}} ) = -3(1+(\frac{1}{2} +...+\frac{1}{2^{11}} ))= -3(1+(2^{10} +...+2+1)/2^{11})...$

Figure out what $2^{10} + 2^9+...+2 + 1$ is and your pretty much done.

I just tried it, still couldn't get it right.

**mr fantastic** · April 1st 2010, 06:13 PM

Originally Posted by dylan5188

Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation. A.

I got -4 which is wrong.
Any thought on that one ?

a = -6, r = -1/2 and n = 13. Substitute into the usual formula. Please show your work and say where you're stuck if you need mor help.

**Anonymous1** · April 1st 2010, 06:16 PM

Originally Posted by dylan5188

I just tried it, still couldn't get it right.

What do you mean? You couldn't figure out 2^10+... thing?

While I have MATLAB open

EDU>> 2^10+2^9+2^8+2^7+2^6+2^5+2^4+2^3+2^2+2+1

ans =

2047

So your answer is -5.9985...

Unless I made an algebra mistake somewhere, but I can't see one.

**dylan5188** · April 1st 2010, 06:16 PM

Originally Posted by mr fantastic

a = -6, r = -1/2 and n = 13. Substitute into the usual formula. Please show your work and say where you're stuck if you need mor help.

yea I got r is -1/2. which has a abs value<1.
and put in to a/(1-r) then I got -4. I didnt use n though

**dylan5188** · April 1st 2010, 06:19 PM

Originally Posted by Anonymous1

What do you mean? You couldn't figure out 2^10+... thing?

I meant I tried your way and got -5.99.... from calculator, but when I submit this answer, is shows answer incorrect.

**mr fantastic** · April 1st 2010, 06:22 PM

Originally Posted by dylan5188

yea I got r is -1/2. which has a abs value<1.
and put in to a/(1-r) then I got -4. I didnt use n though

You were told all the way back in post #2 that you had the wrong formula!

Your series is NOT infinite, it is FINITE. Go to your class notes or textbook, find the formula for a finite geometric series and substitute the appropriate values into it.

**Anonymous1** · April 1st 2010, 06:25 PM

OOO its alternating oops.

Try -4.0005

**dylan5188** · April 1st 2010, 06:32 PM

Originally Posted by mr fantastic

You were told all the way back in post #2 that you had the wrong formula!

Your series is NOT infinite, it is FINITE. Go to your class notes or textbook, find the formula for a finite geometric series and substitute the appropriate values into it.

Finally got it. I didnt notice is finite, the whole chapter is about infinite.
Thanks

**dylan5188** · April 1st 2010, 06:33 PM

Originally Posted by Anonymous1

OOO its alternating oops.

Try -4.0005

Yea that's the right answer.
Thanks

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