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.
thanks in advance
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?
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?
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).
The formula you used above is good, and in case you're interested here's another way of looking at it:
Consider the series
and the sequence of its partial sums
Notice that
So we reach the same conclusion that
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
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.