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# Sub numbers and geometric series.

(See attached file for full problem description).

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(Qu. 1)

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[1] = (-2)^1 + 5
= 3

Let n = 2, then a[2] = (-2)^2 + 5
= 9

Similarly, you can work out that a[3] = -3

We recall that the nth partial sum, S[n] = a[1] + a[2] + ... + a[n]
So, the first partial sum:

S[1] = a[1] = 3

The third partial sum:

S[3] = a[1] + a[2] + a[3]
= 3 + 9 + -3
= 9

(Qu. 2)

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)
n-> inf

In question 2 we have:

a = 2/3 and r = -2/3

=> S = (2/3) / (1 - (-2/3))
= 2/5

(Qu. 3)

We can divide this question in two ...

#### Solution Summary

This solution is comprised of a detailed explanation to answer sub numbers and geometric series.

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