Do you mean
a_n = 1 - 4(1/3)^(n-1)
?
If so, then the sequence is
1 - 4*1, 1 - 4/3, 1 - 4/9, 1 - 4/27 ...
= -3, -1/3, 5/9, 23/27, ...
But that's not a geometric sequence. I suspect you meant
a = -4
r = 1/3
If so, then that sequence is
-4, -4/3, -4/9, -4/27, ...
And like all GP, S = 1/(1-r) = -4/(1 - 1/3) = -4/(2/3) = -4 * 3/2 = -6
If I still didn't parse your text correctly, then I assume you can fix it and work it out using the methods above.
consider the infinite geometric series n=1 -4(1/3)^n-1 .
i need help with writing the first four terms of the series and finding the sum if it has a sum.
5 answers
no, the problem that i have has Σ with an infinite sign and n = 1. after it it has -4(1/3)^n-1
then that is what I did in my 2nd half above.
∞
Σ -4(1/3)^(n-1) = -4 (1 + 1/3 + 1/9 + 1/27) ... = -6
n=1
I see I did make a typo: S∞ = a/(1-r)
not 1/(1-r)
Looks like it's time for a review of your textbook.
∞
Σ -4(1/3)^(n-1) = -4 (1 + 1/3 + 1/9 + 1/27) ... = -6
n=1
I see I did make a typo: S∞ = a/(1-r)
not 1/(1-r)
Looks like it's time for a review of your textbook.
fkn idiot... we don't receive textbooks
oobleck Thank you so much for the answer
1 answer