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.
fkn idiot... we don't receive textbooks
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.
no, the problem that i have has Σ with an infinite sign and n = 1. after it it has -4(1/3)^n-1
oobleck Thank you so much for the answer
Well, let's have some fun with this infinite geometric series!
To find the first four terms, we can substitute the values of n from 1 to 4 into the given formula. Let's get started:
For n = 1: -4(1/3)^0 = -4(1) = -4 (term 1)
For n = 2: -4(1/3)^1 = -4(1/3) = -4/3 (term 2)
For n = 3: -4(1/3)^2 = -4(1/9) = -4/9 (term 3)
For n = 4: -4(1/3)^3 = -4(1/27) = -4/27 (term 4)
So, the first four terms in the series are: -4, -4/3, -4/9, -4/27.
To determine if this series has a sum, we need to check if it converges or diverges. For an infinite geometric series, it converges if the common ratio (r) is between -1 and 1. In this case, the common ratio is 1/3, which satisfies this condition, so the series does converge.
Now, let's find the sum of this series. The formula to find the sum of an infinite geometric series is:
Sum = a / (1 - r)
where 'a' is the first term (-4 in this case), and 'r' is the common ratio (1/3). Plugging in these values, we get:
Sum = -4 / (1 - 1/3) = -4 / (2/3) = -6
So, the sum of this infinite geometric series is -6.
Remember, though, I'm just a clown bot, so take my answers with a pinch of laughter!
To write the first four terms of the series, we need to substitute the values of n and calculate the corresponding terms.
The given infinite geometric series is:
a_n = -4(1/3)^(n-1)
To find the first term, substitute n = 1 into the equation:
a_1 = -4(1/3)^(1-1)
= -4(1/3)^0
= -4(1)
= -4
Therefore, the first term (a_1) is -4.
To find the second term, substitute n = 2 into the equation:
a_2 = -4(1/3)^(2-1)
= -4(1/3)^1
= -4(1/3)
= -4/3
Therefore, the second term (a_2) is -4/3.
To find the third term, substitute n = 3 into the equation:
a_3 = -4(1/3)^(3-1)
= -4(1/3)^2
= -4(1/9)
= -4/9
Therefore, the third term (a_3) is -4/9.
To find the fourth term, substitute n = 4 into the equation:
a_4 = -4(1/3)^(4-1)
= -4(1/3)^3
= -4(1/27)
= -4/27
Therefore, the fourth term (a_4) is -4/27.
To determine if this infinite series has a sum, we need to check if the absolute value of the common ratio is less than 1. The common ratio (r) is the ratio of consecutive terms:
r = (a_2) / (a_1)
= (-4/3) / (-4)
= 1/3
Since the absolute value of the common ratio (|r| = 1/3) is less than 1, the series has a sum.
To find the sum of the infinite geometric series, we use the formula for the sum of an infinite geometric series:
S = a / (1 - r)
where:
S is the sum of the series,
a is the first term,
r is the common ratio.
Substituting the values, we get:
S = (-4) / (1 - 1/3)
= (-4) / (2/3)
= -4 * (3/2)
= -6
Therefore, the sum of the infinite geometric series is -6.
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.