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Consider the series:
1+1/2+1/4+1/8....
How many terms of this series must be taken before the sum first exceeds 1999/1000?
What I did was first input 1999/1000 into the series equation, so
Sn= n/2 (2u1 + (n-1)d)
1999/1000 = n/2 (2 + (n-1)(1/2) )
1999/1000 = n (2+ (1/2)n - (1/2) )
0 = 1/2 n^2 +3/2 n - 1999/500
Then i used the quad formula and got
1.7, -4.7
... but its wrong lol =|
i'm not sure what im supposed to do
You are using the sum of an arithmetic sequence on a geometric sequence. Try doing the same but usingQuote:
Originally Posted by juliak
Where a is the first term and r is the common ratio
Well, the series is a geometric progression and not arithmetic progression. Hence ur formula for sum is wrong.
The formula for sum of a geometric progression is:
where a is in the initial term in the series
r is the ratio of the second term to first term or third term to second term and so on....(in this case it is 1/2)
n is the no. of terms reqd
The denominator depends on whether r>1 or r<1. In this case, r<1, hence the formula
Substituting for above parameters and taking the fact that it is just less than 1999/1000, u get,
Solve for n.
thank you!
I simplified down to
1.9995 < 1/2 ^n
how can i find the n from that
Quote:
Originally Posted by juliak
Note I said that