(See attached file for full problem description).
We are given the expression for the nth term of the sequence a[n]:
a[n] = (-2)^n + 5
So we can find any term we like by substituting in our chosen value for n.
Let n = 1, then a = (-2)^1 + 5
Let n = 2, then a = (-2)^2 + 5
Similarly, you can work out that a = -3
We recall that the nth partial sum, S[n] = a + a + ... + a[n]
So, the first partial sum:
S = a = 3
The third partial sum:
S = a + a + a
= 3 + 9 + -3
With both arithmetic and geometric series it can be quite easy to make "out by one" errors, so let's clarify the definition first:
A geometric series has the form:
a + a.r + a.r^2 + ... + a.r^(n-1) + ...
The nth term in the series, a[n] = a.r^(n-1)
We will need the nth partial sum, S[n], which we can derive as follows.
S[n] = a + a.r + a.r^2 + ... + a.r^(n-1)
r.S[n] = a.r + a.r^2 + ... + a.r^(n-1) + a.r^n
(1-r).S[n] = a - a.r^n
S[n] = a.(1 - r^n)/(1 - r)
As in the textbook. Now you can see if -1 < r < 1, then as n tends to infinity, r^n will fall ever closer to zero and we have:
Lim S[n] = a / (1 - r)
In question 2 we have:
a = 2/3 and r = -2/3
=> S = (2/3) / (1 - (-2/3))
We can divide this question in two ...
This solution is comprised of a detailed explanation to answer sub numbers and geometric series.