4.8 Q-11 | Sequence And Series | Mathematics 11

Exercise 4.8 — Question 11

Topic: Arithmetic-Geometric Series and Sigma Notation Sums

Note: The original question statement for Q11 was not provided. The solution framework below covers the standard techniques tested under SLOs M-11-A-27 and M-11-A-26 for Exercise 4.8-type problems. Please insert the exact question text when available.

Key Formulas

Arithmetic-Geometric Sequence

An arithmetic-geometric sequence is formed by multiplying corresponding terms of an arithmetic sequence and a geometric sequence.

If the arithmetic sequence has first term $a$ and common difference $d$ , and the geometric sequence has first term $1$ and common ratio $r$ , then:

$T_{n} = [a + (n - 1) d] r^{n - 1}$

Sum to $n$ Terms

Using the subtraction method (multiply $S_{n}$ by $r$ and subtract):

$S_{n} = \frac{a - [ a + ( n - 1 ) d ] r ^{n}}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}, r \neq = 1$

Sum to Infinity

When $∣ r ∣ < 1$ , as $n \to \infty$ , $r^{n} \to 0$ :

$S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$

Sigma Notation Sums (SLO M-11-A-26)

Sum	Sigma Notation	Closed Form
First $n$ natural numbers	$k = 1 \sum n k$	$\frac{n ( n + 1 )}{2}$
Squares of first $n$ natural numbers	$k = 1 \sum n k^{2}$	$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
Cubes of first $n$ natural numbers	$k = 1 \sum n k^{3}$	$[\frac{n ( n + 1 )}{2}]^{2}$

Worked Example

Find the sum to infinity of the series: $1 + 2 \cdot \frac{1}{2} + 3 \cdot (\frac{1}{2})^{2} + 4 \cdot (\frac{1}{2})^{3} + \dots$

Identify: $a = 1$ , $d = 1$ , $r = \frac{1}{2}$ , and $∣ r ∣ < 1$ so the series converges.

Apply the formula:

$S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$

$S_{\infty} = \frac{1}{1 - \frac{1}{2}} + \frac{1 \cdot \frac{1}{2}}{( 1 - \frac{1}{2} ) ^{2}}$

$S_{\infty} = \frac{1}{\frac{1}{2}} + \frac{\frac{1}{2}}{\frac{1}{4}} = 2 + 2 = 4$

Method: Subtraction Technique for Arithmetic-Geometric Series

Let $S = a + (a + d) r + (a + 2 d) r^{2} + \dots$

Step 1: Write $S$ : $S = a + (a + d) r + (a + 2 d) r^{2} + (a + 3 d) r^{3} + \dots$

Step 2: Multiply both sides by $r$ : $r S = a r + (a + d) r^{2} + (a + 2 d) r^{3} + \dots$

Step 3: Subtract $r S$ from $S$ : $S (1 - r) = a + d r + d r^{2} + d r^{3} + \dots = a + \frac{d r}{1 - r}$

Step 4: Solve for $S$ : $S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}, ∣ r ∣ < 1$

Exercise 4.8 — Question 11

Key Formulas

An arithmetic-geometric sequence is formed by multiplying corresponding terms of an arithmetic sequence and a geometric sequence.

If the arithmetic sequence has first term

a

and common difference

d

, and the geometric sequence has first term

1

and common ratio

r

, then:

T_{n} = [a + (n - 1) d] r^{n - 1}

Using the subtraction method (multiply

S_{n}

r

and subtract):

S_{n} = \frac{a - [ a + ( n - 1 ) d ] r ^{n}}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}, r \neq = 1

When

∣ r ∣ < 1

, as

n \to \infty

r^{n} \to 0

S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}

Sum

Sigma Notation

Closed Form

First

n

natural numbers

k = 1 \sum n k

\frac{n ( n + 1 )}{2}

Squares of first

n

natural numbers

k = 1 \sum n k^{2}

\frac{n ( n + 1 ) ( 2 n + 1 )}{6}

Cubes of first

n

natural numbers

k = 1 \sum n k^{3}

[\frac{n ( n + 1 )}{2}]^{2}