IB Mathematics AI AHL The sum of infinite geometric sequences Study Notes - New Syllabus

Solution:

First term: $a = \frac{1}{2}$

Common ratio:
$ r = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} $

A geometric series has a sum to infinity only if $|r| < 1$.

Since $r = \frac{1}{2}$ and $|r| < 1$, the sum does exist.

$
S_\infty = \frac{a}{1 – r}
$

$
S_\infty = \frac{\frac{1}{2}}{1 – \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1
$

Solution:

(i)
Let
$
x = 0.333\ldots
$

Multiply both sides by 10

$
10x = 3.333\ldots
$

Subtract the original equation:
$
10x – x = 3.333\ldots – 0.333\ldots
\Rightarrow 9x = 3
\Rightarrow x = \frac{3}{9} = \frac{1}{3}
$

(ii)
$
x = 0.999\ldots
$
Multiply both sides by 10:

$
10x = 9.999\ldots
$

Subtract the original equation
$
10x – x = 9.999\ldots – 0.999\ldots
\Rightarrow 9x = 9
\Rightarrow x = 1
$

Solution:

Initial Fall (Downward Motion):

The ball first falls $64\, \text{m}$ straight down.

After the first impact:

It rises to $\dfrac{3}{4} \times 64 = 48\, \text{m}$ (upward motion),
Then falls $48\, \text{m}$ (downward motion),
Then rebounds to $\dfrac{3}{4} \times 48 = 36\, \text{m}$, etc.

This forms two infinite geometric series:

One for downward motion (excluding the first fall),
One for upward motion.

Total Downward Motion:

We include the first fall of 64 m and then the infinite series:

$
\text{Downward distance} = 64 + 48 + 36 + 27 + \dots
$

This is a geometric series with:

First term $a = 48$,
Common ratio $r = \dfrac{3}{4}$.

$
\text{Sum of infinite downward motion} = 64 + \sum_{n=1}^{\infty} 48\left( \dfrac{3}{4} \right)^{n-1}
= 64 + \frac{48}{1 – \frac{3}{4}} = 64 + \frac{48}{\frac{1}{4}} = 64 + 192 = 256\, \text{m}
$

Total Upward Motion:

$
\text{Upward distance} = 48 + 36 + 27 + \dots
$

This is again a geometric series with:

First term $a = 48$,
Common ratio $r = \dfrac{3}{4}$,

$
\text{Sum} = \frac{48}{1 – \frac{3}{4}} = \frac{48}{\frac{1}{4}} = 192\, \text{m}
$

Total Vertical Distance:

$
\text{Total distance} = \text{Total downward} + \text{Total upward}
= 256 + 192 = \rm{448\, \text{m}}
$

Solution:

This is a geometric series with \( u_1 = 1 \) and \( r = 2 \). Since \( |r| \geq 1 \), the series diverges. The partial sums grow infinitely:
\( S_n = 2^n – 1 \rightarrow \infty \quad \text{as } n \rightarrow \infty \)