4.6 Q-3 | Sequence And Series | Mathematics 11

Exercise 4.6 — Question 3

Note: The specific problem statement for Exercise 4.6 Q3 should be inserted here from the FBISE textbook. The worked solution below follows the standard format for summation/series problems covered in this exercise.

Key Summation Formulas

For problems involving sums of natural numbers, squares, and cubes:

$\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$

$\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

$\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

Arithmetic-Geometric Series

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:

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

Sum to $n$ terms (using the subtraction method, $r \neq = 1$ ): $S_{n} = \frac{a - [ a + ( n - 1 ) d ] r ^{n}}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}$

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

Worked Solution

(Insert the specific question from the FBISE textbook here and apply the appropriate formula from above.)

Step 1: Identify whether the series involves standard summation formulas (∑n, ∑n², ∑n³) or an arithmetic-geometric structure.

Step 2: Apply the relevant formula.

Step 3: Simplify algebraically to obtain the final answer.

Exercise 4.6 — Question 3

Key Summation Formulas

For problems involving sums of natural numbers, squares, and cubes:

$\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}$

$\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

$\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

Arithmetic-Geometric Series

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:

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

Sum to $n$ terms (using the subtraction method, $r \neq = 1$ ): $S_{n} = \frac{a - [ a + ( n - 1 ) d ] r ^{n}}{1 - r} + \frac{d r ( 1 - r ^{n - 1} )}{( 1 - r ) ^{2}}$

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

Worked Solution

(Insert the specific question from the FBISE textbook here and apply the appropriate formula from above.)

Step 1: Identify whether the series involves standard summation formulas (∑n, ∑n², ∑n³) or an arithmetic-geometric structure.

Step 2: Apply the relevant formula.

Step 3: Simplify algebraically to obtain the final answer.