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[h] IB Mathematics AA HL Flashcards- The sum of an infinite geometric sequence
[q] Infinite Geometric Sequence
[a]
1// In 1.3, we started to look at how to sum the terms of a geometric sequence. However, this was only for a finite number of terms.
In this section, we will look at when it is possible to sum an infinite number of terms, and how to evaluate the value of the sum if we indeed can. Clearly, if we have an increasing sequence, such as: \(2, 6, 18, 54, 156, \dots\), the sum will tend to infinity, or diverge.
[q]
But what if the terms are decreasing? You can probably see that \(1 + 0.01 + 0.0001 + \dots\) will converge to something less than 2, but you may also know the simple harmonic series: \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots\) does NOT converge to any finite value. So, what about geometric series?
[a]
CONVERGENCE// Take \(\sum_{k=0}^{n} U_1 r^k = U_1 + U_1 r + U_1 r^2 + U_1 r^3 + \dots\), and write it as:
\[
\lim_{n \to \infty} \left(U_1 + U_1 r + U_1 r^2 + \dots + U_1 r^n\right) = U_1 \times \lim_{n \to \infty} \left(1 + r + r^2 + \dots + r^n\right) \times \left(\frac{1-r^{n+1}}{1-r}\right)
\]
[q]
As \( \lim_{n \to \infty} r^{n+1} = 0 \) as \( n \to \infty \), we have \( U_1 \times \lim_{n \to \infty} \left(\frac{1}{1-r}\right) \) when \( |r| < 1 \).
Then, as \( U_1 \) and \( r \) are constant, \( U_1 \times \lim_{n \to \infty} \left(\frac{1}{1-r}\right) = \frac{U_1}{1-r} \). So, we are given:
\[
S_\infty = \frac{U_1}{1-r}, \quad \text{when} \, |r| < 1 \, \text{(between -1 \& 1)}
\]
[a]
E.G.1 // Does \(-2 + 5/2 + 25/8 + 125/32 + \dots\) converge?
Find \( r \):
\[
r = \frac{U_2}{U_1} = \frac{5/2}{-2} = \frac{5}{4} = -1.25
\]
Since \(|-1.25| > 1\), it diverges.
[q]
E.G.2 // Find the sum of \(1 + \frac{4}{3} + \frac{4^2}{9} + \frac{4^3}{27} + \dots\)
Find \( r \) by doing \( \frac{U_{n+1}}{U_n} \), e.g.,
\[
U_2 = \frac{4}{3}, \quad \frac{U_2}{U_1} = \frac{4/3}{1} = \frac{4}{3}
\]
[a]
Use formula:
\[
S_\infty = \frac{U_1}{1-r} = \frac{3}{1-\frac{4}{3}} = 3
\]
We observe that \( S_3 = 2.1 \), \( S_6 = 2.74 \), \( S_{20} \approx 2.999 \). So, this seems correct.
[q]
E.G.3 // If \( S_3 = 62.755 \) and \( S_\infty = 440 \), find the common ratio:
Gather all info:
\[
S_3 = \frac{U_1(1-r^3)}{1-r} = 62.755, \quad S_\infty = \frac{U_1}{1-r} = 440
\]
[a]
This is a system of 2 equations, with 2 unknowns. Using substitution:
\[
\frac{U_1}{1-r} = 440 \quad \Rightarrow \quad U_1 = 440(1-r)
\]
Sub into \( S_3 \) equation:
\[
\frac{440(1-r)(1-r^3)}{1-r} = 62.755 \quad \Rightarrow \quad 440r – 440r^3 = 62.755 \quad \Rightarrow \quad r^3 = 0.857375 \quad \Rightarrow \quad r \approx 0.95
\]
[a] Standard form:
A number in the format , \( \pm a \times10^k \) , where \(1\leqslant a\leqslant 10 \) and \(k\in Z\) (k is an integer)
[q] Standard Form – Solved Examples 1
The diameter of a spherical planet is 6 × 104 km .
(a) Write down the radius of the planet. [1]
The volume of the planet can be expressed in the form π(a × 10k) km3 where 1 ≤ a <10 and k ∈ \(\mathbb{Z}\)
(b) Find the value of a and the value of k .
[a] Answer – Solved Examples 1
(a) \(3 \times 10^4\)
(b) \(\frac{4}{3}\pi(3 \times 10^4)^3\)
\(=\frac{4}{3}\pi\times 27 \times 10^{12}\)
\(=\pi(3.6 \times 10^{13})(km)^3\)
Hence
\(a= 3.6 \) , \(k =13\)
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