math extra credit?

Question

If the first two partial sums in a geometric series are 1 and -1, respectively, what is the third partial sum?

Answer 1

first term = a =1
second term = ar
third term = ar^2

Partial Sums
a = 1
a +ar = -1

Put a = 1 into a + ar = -1

1 + r = -1
r = -2

So the third partial sum (a + ar + ar^2) = 3

Answer 2

Where did you get this question?

You have a problem here ... there are an infinite number of solutions to the question. You cannot determine a unique solution to a series with only two given points. Especially for a geometric series. (e.g. a polynomial equation in the nth degree) To solve a series, you need at least 3 points, often more. For example, the series 2, 3 ... may be followed with the following sets:
a. ...4, 5, 6 ... {integers or n}
b. ...5, 7, 11 ... {prime numbers}
c. ...5, 8, 13 ... {2n-1}
and many more possibilities. And that is for linear equations.

Now, for geometric ... your series is the sum (Greek letter upper case sigma, which I cannot produce here) from i=1 to n of some yet undetermined polynomial. The first partial sum, S(i=1) or S(1), is where you let i=1. The second partial sum, S(i=2) or S(2), is the sum of the polynomial's result for i=1, plus the polynomial's result for i=2. Going on, for the third partial sum, S(i=3) or S(3), you want the sum of the polynomial's result for i=1, plus the polynomial's result for i=2, plus the polynomial's result for i=3.

Now, the trick here is that you want to determine a unique polynomial with only two data points.

What we have from the given data is S(1) = 1 and S(2) = -1. The unknown polynomial, P, must then equal 1 for i=1, or P(i=1) = P(1) = 1, and P(i=2) = P(2) = S(2) - S(1) = -1 - 1 = -2.

So, now we need to see if there is something which forces a polynomial to be unique for this criteria: P(1) = 1 and P(2) = -2. The answer is: unfortunately not, for many polynomials include these two points.

As just one example, take the equation P(i) = -2^(i - 1). For P(1) = -2^(1-1) = -2^0 = 1. P(2) = -2^(2-1) = -2^1 = -2. P(3) = -2^(3-1) = -2^2 = 4. For this, S(3) = P(1) + P(2) +P(3) = 1 +(-1) + 4 = 4.
But again, other solutions can be found with differing P(3) and, therefore, differing S(3) results.

Sorry, there is no unique solution.

Answer 3

Let's get this straight . . . we give you the answer and you get extra credit. Hm-m-m-m. Did you type the question by yourself, or did someone else do it for you?

Answer 4

sequence is a, ar, ar², ......

S1 = a = 1
S2 = a + ar = -1
so 1 + r = -1
thus r = -2
Hence ar² = 4

So S3 = S2 + ar² = 3

Sequence is 1, -2, 4, -8, ....., (-2)^(n - 1), .....