# Math Help - a series problems

1. ## a series problems

Hi would appreciate some help in how i can solve this problem..

i have to find the minimum number of terms which would make the sum of the series 0.121+0.242+0.484 ...
to become greater than 80.

2. Originally Posted by zzizi
Hi would appreciate some help in how i can solve this problem..

i have to find the minimum number of terms which would make the sum of the series 0.121+0.242+0.484 ...
to become greater than 80.

Go to the formula for the sum of a geometric series. Substitute that the first term is 0.121 and the common ratio is 2. Put it greater than 80 and solve the resulting inequality for the minimum value of n.

3. MrFantastic thank you for the reply.

Pls correct me if this is wrong...

should i do this?

S3 = 0.121(1-2^3)/ (1-2)>80

4. Originally Posted by zzizi
MrFantastic thank you for the reply.

Pls correct me if this is wrong...

should i do this?

S3 = 0.121(1-2^3)/ (1-2)>80
Please post the formula for the sum of a geometric series and then show how you used it to get the above.

5. ok... this is the formula i just used
S(subN) =a(1-r^n)/1-r

and then i set it greater than 80

should i use this formula or is there another one?

6. Originally Posted by zzizi
ok... this is the formula i just used
S(subN) =a(1-r^n)/1-r

and then i set it greater than 80

should i use this formula or is there another one?
Yes. So 0.121(1-2^3)/ (1-2)>80 (post #3) should be 0.121(1-2^n)/ (1-2)>80 and you solve for the minimum integer value of n.

7. Oh right thanks.... so then the inequality would require me to use logs to solve? ... i will post my working out in a min...

8. i cant type up all the symbols... but what i tried to do was the following:
0.121-2^n/1-2 >80
then i subtracted 2 from the denominator which left -1 .. i then divided both sides by neg 1 where switched signs to this
0.121-2^n<-80
then added 0.121 to both sides
-2^n<-79.9
i then to log of both side
n(ln)-2<(ln)-79.9
to get this
n< (ln) -79.9/ (ln) -2
therefore
n<6.3
pls tell me this right?

anyone?

9. Originally Posted by zzizi
i cant type up all the symbols... but what i tried to do was the following:
0.121-2^n/1-2 >80
then i subtracted 2 from the denominator which left -1 .. i then divided both sides by neg 1 where switched signs to this
0.121-2^n<-80
then added 0.121 to both sides
-2^n<-79.9
i then to log of both side
n(ln)-2<(ln)-79.9
to get this
n< (ln) -79.9/ (ln) -2
therefore
n<6.3
pls tell me this right?

anyone?
Substitute your answer into the inequality. Does it work? Note: n is meant to be an integer. And IF your answer is correct, it should read n > 6.3 ....

Personally, I'd solve the inequality using trial-and-error.

10. yes i think it does work.. because i tried adding up 6 terms and it does sum to over 80... i need to brush up on this area.. thanks very much for help.

11. Originally Posted by zzizi
yes i think it does work.. because i tried adding up 6 terms and it does sum to over 80
[snip]
Really? Please show the details ....

12. oh.... erm... ok now im alittle confused... what i meant was that i multiplied 0.121 six times by 2 in the calculator and it added up to over 80.
pls do shed some light as to how i should find the answer and was my working out the right way to solve this problem?

13. Originally Posted by zzizi
oh.... erm... ok now im alittle confused... what i meant was that i multiplied 0.121 six times by 2 in the calculator and it added up to over 80.
[snip]
Since 2^6 = 64 I doubt very much that 0.121 multiplied six times by 2 will be greater than 80.

I have told you how to do the question, but until you can do the basic arithmetic correctly (and I'm not going to that for you) you will not get the correct answer.

(It is simple enough to write out and then add up the first six terms - doing so should have quickly shown you that your answer was wrong).

14. ok point taken... thanks

15. Originally Posted by zzizi
Hi would appreciate some help in how i can solve this problem..

i have to find the minimum number of terms which would make the sum of the series 0.121+0.242+0.484 ...
to become greater than 80.

The formula you used above is good, and in case you're interested here's another way of looking at it:

$0.121+0.242+0.484+...$

$=0.121(1+2+4+8+...)$

Consider the series

$1+2+4+8+...$

and the sequence of its partial sums

$\{a_n\}=1,3,7,15,31,...$

Notice that $a_n=2^n-1$

So we reach the same conclusion that $0.121(2^n-1)>80$

Like mr fantastic mentioned, trial and error is pretty easy here, especially if you are familiar with powers of 2 since they come up a lot.

But if you feel you absolutely must solve the inequality exactly, then I can show you the first couple steps

$0.121\left(2^n-1\right)>80$

$2^n-1>\frac{80}{0.121}$

$2^n>\frac{80}{0.121}+1$

You will end up using logarithms.. well I practically did the problem for you already. Anyway good luck. By the way there is a forum that discusses math formatting using LaTeX.

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