4.8 Q-3 | Sequence And Series | Mathematics 11

Exercise 4.8 — Question 3

Topic: Arithmetic-Geometric Series and Summation Formulas

This question involves applying the summation formulas for $\sum n$ , $\sum n^{2}$ , $\sum n^{3}$ and/or working with arithmetic-geometric sequences and series.

Key Formulas

Sum of first $n$ natural numbers: $\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$

Sum of squares of first $n$ natural numbers: $\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

Sum of cubes of first $n$ natural numbers: $\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

General term of an arithmetic-geometric sequence:

If ${a_{n}}$ is arithmetic with first term $a$ and common difference $d$ , and ${b_{n}}$ is geometric with first term $1$ and common ratio $r$ , then the arithmetic-geometric sequence has general term: $T_{n} = [a + (n - 1) d] r^{n - 1}$

Sum to $n$ terms of an arithmetic-geometric series: $S_{n} = \frac{a - [ a + ( n - 1 ) d ] r ^{n}}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}, r \neq = 1$

Sum to infinity (when $∣ r ∣ < 1$ ): $S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$

Worked Solution

⚠️ Note: The specific problem statement for Q3 of Exercise 4.8 is not available in this note. Please add the question text here. The solution method below follows the standard approach for this exercise type.

Step 1: Identify whether the series is a pure summation ( $\sum n$ , $\sum n^{2}$ , $\sum n^{3}$ ) or an arithmetic-geometric series.

Step 2: Write the general term $T_{n}$ of the series.

Step 3: Apply the appropriate formula:

For $\sum n$ , $\sum n^{2}$ , $\sum n^{3}$ : use the closed-form formulas above.
For arithmetic-geometric series: use the subtraction method or the formula for $S_{n}$ .

Step 4: Simplify and state the final answer.

Example (Arithmetic-Geometric Series)

Find the sum to $n$ terms of: $1 + 2 r + 3 r^{2} + 4 r^{3} + \dots$

General term: $T_{k} = k \cdot r^{k - 1}$

Using the subtraction method:

Let $S_{n} = 1 + 2 r + 3 r^{2} + \dots + n r^{n - 1}$

$r S_{n} = r + 2 r^{2} + 3 r^{3} + \dots + n r^{n}$

Subtracting: $S_{n} (1 - r) = 1 + r + r^{2} + \dots + r^{n - 1} - n r^{n} = \frac{1 - r ^{n}}{1 - r} - n r^{n}$

$S_{n} = \frac{1 - r ^{n}}{( 1 - r ) ^{2}} - \frac{n r ^{n}}{1 - r}, r \neq = 1$

For $∣ r ∣ < 1$ , as $n \to \infty$ : $S_{\infty} = \frac{1}{( 1 - r ) ^{2}}$

Exercise 4.8 — Question 3

Topic: Arithmetic-Geometric Series and Summation Formulas

This question involves applying the summation formulas for $\sum n$ , $\sum n^{2}$ , $\sum n^{3}$ and/or working with arithmetic-geometric sequences and series.

Key Formulas

Sum of first $n$ natural numbers: $\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$

Sum of squares of first $n$ natural numbers: $\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

Sum of cubes of first $n$ natural numbers: $\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

General term of an arithmetic-geometric sequence:

Sum to $n$ terms of an arithmetic-geometric series: $S_{n} = \frac{a - [ a + ( n - 1 ) d ] r ^{n}}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}, r \neq = 1$

Sum to infinity (when $∣ r ∣ < 1$ ): $S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}$

Worked Solution

Step 1: Identify whether the series is a pure summation ( $\sum n$ , $\sum n^{2}$ , $\sum n^{3}$ ) or an arithmetic-geometric series.

Step 2: Write the general term $T_{n}$ of the series.

Step 3: Apply the appropriate formula:

For $\sum n$ , $\sum n^{2}$ , $\sum n^{3}$ : use the closed-form formulas above.
For arithmetic-geometric series: use the subtraction method or the formula for $S_{n}$ .

Step 4: Simplify and state the final answer.

Example (Arithmetic-Geometric Series)

Find the sum to $n$ terms of: $1 + 2 r + 3 r^{2} + 4 r^{3} + \dots$

General term: $T_{k} = k \cdot r^{k - 1}$

Using the subtraction method:

Let $S_{n} = 1 + 2 r + 3 r^{2} + \dots + n r^{n - 1}$

$r S_{n} = r + 2 r^{2} + 3 r^{3} + \dots + n r^{n}$

Subtracting: $S_{n} (1 - r) = 1 + r + r^{2} + \dots + r^{n - 1} - n r^{n} = \frac{1 - r ^{n}}{1 - r} - n r^{n}$

$S_{n} = \frac{1 - r ^{n}}{( 1 - r ) ^{2}} - \frac{n r ^{n}}{1 - r}, r \neq = 1$

For $∣ r ∣ < 1$ , as $n \to \infty$ : $S_{\infty} = \frac{1}{( 1 - r ) ^{2}}$