Prove, by induction, the formula for the sum of the first n terms of a geometric series. That is, prove that, for r ≠ 1: a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1-r^n)}{1-r}. (b) By writing the recurring part as an infinite geometric series, express the following number as a fraction of integers: 5\cdot 2i = 5.2121212121...

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Prove, by induction, the formula for the sum of the first n terms of a geometric series - Leaving Cert Mathematics - Question 4 - 2012

Question 4

Prove, by induction, the formula for the sum of the first n terms of a geometric series. That is, prove that, for r ≠ 1: a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1... show full transcript

Worked Solution & Example Answer:Prove, by induction, the formula for the sum of the first n terms of a geometric series - Leaving Cert Mathematics - Question 4 - 2012

Step 1

Prove, by induction, the formula for the sum of the first n terms of a geometric series.

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Answer

We start by defining the series:

Let P(n):

$S_n = a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1-r^n)}{1-r}$

Base Case (n = 1): Check P(1):

$S_1 = a = \frac{a(1-r^1)}{1-r} = a$

This is true.

Inductive Step: Assume P(k) holds, meaning:

$S_k = a + ar + ar^2 + ... + ar^{k-1} = \frac{a(1-r^k)}{1-r}$

Then, we must show P(k + 1):

$S_{k+1} = S_k + ar^{k}$

Substituting the assumption:

$S_{k+1} = \frac{a(1-r^k)}{1-r} + ar^{k}$

Combine the terms:

$= \frac{a(1-r^k) + ar^{k}(1-r)}{1-r}$ $= \frac{a(1-r^k + r^k - r^{k+1})}{1-r}$ $= \frac{a(1-r^{k+1})}{1-r}$

This completes the inductive step, proving that P(n) holds for all n ∈ ℕ.

Step 2

By writing the recurring part as an infinite geometric series, express the following number as a fraction of integers:

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Answer

To express $5\cdot 2i$ as described, we start with:

$5\cdot 2i = 5.21212121...$

We rewrite this as:

$= 5 + 0.21212121...$

Now express the recurring part as an infinite geometric series:

Let x = 0.21212121...

The formula for a recurring geometric series is:

$x = \frac{a}{1 - r}$

where a is the first term and r is the common ratio. In this case:

a = 21/100,

and r = 1/100.

So, we can calculate:

$x = \frac{21/100}{1 - 1/100} = \frac{21/100}{99/100} = \frac{21}{99}$

Then, we substitute back:

$5 \cdot 2i = 5 + \frac{21}{99} = \frac{5 \cdot 99 + 21}{99} = \frac{495 + 21}{99} = \frac{516}{99}$

Simplifying yields:

$5\cdot 2i = \frac{172}{33}$ .

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Prove, by induction, the formula for the sum of the first n terms of a geometric series - Leaving Cert Mathematics - Question 4 - 2012

Worked Solution & Example Answer:Prove, by induction, the formula for the sum of the first n terms of a geometric series - Leaving Cert Mathematics - Question 4 - 2012

Prove, by induction, the formula for the sum of the first n terms of a geometric series.

By writing the recurring part as an infinite geometric series, express the following number as a fraction of integers:

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