1. ## infinite series problem

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 ?

2. 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.

3. 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.

4. $-6+3-3(\frac{1}{2}+...+\frac{1}{2^{11}} ) = -3(1+(\frac{1}{2} +...+\frac{1}{2^{11}} ))= -3(1+(2^{10} +...+2+1)/2^{11})...$

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

5. 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.

6. 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.

7. 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

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

8. 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

9. 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.

10. 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.

11. OOO its alternating oops.

Try -4.0005

12. 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

13. Originally Posted by Anonymous1
OOO its alternating oops.

Try -4.0005