Question

Find the sum to \(n\) terms of the series \(1 \cdot 3 \cdot 5+3 \cdot 5 \cdot 7+5 \cdot 7 \cdot 9+\ldots\) is : (a) \(n\left(8 n^{3}+11 n^{2}-n+3\right)\) (c) \(B \pi^{3}+12 n^{2}-2 n-3\) (b) \(n\left(2 n^{3}+8 n^{2}+7 n-2\right)\) (d) none of these

Short answer

Given series: \(1 \cdot 3 \cdot 5, 3 \cdot 5 \cdot 7, 5 \cdot 7 \cdot 9, \ldots\) Options: (a) \(n(8 n^{3}+11 n^{2}-n+3)\) (b) \(n\left(2 n^{3}+8 n^{2}+7 n-2\right)\) (c) \(B \pi^{3}+12 n^{2}-2 n-3\) (d) none of these Answer: (a) \(S_n = n(8n^3 + 11n^2 - n + 3)\)

Step-by-step solution

Identifying the pattern of the series

List the first few terms of the series for easy identification of the pattern: \(1 \cdot 3 \cdot 5, 3 \cdot 5 \cdot 7, 5 \cdot 7 \cdot 9, \ldots\) We observe that the series consists of consecutive odd numbers in triplets.

Writing the general term of the series

For the nth term, we should have three consecutive odd numbers. The nth term should begin with the (2n - 1)th odd number (2n - 1, 2n + 1, 2n + 3) as the product. Therefore, the general term of the series can be written as: \(T_n = (2n - 1)(2n + 1)(2n + 3)\)

Finding the sum of the series up to n terms

To find the sum of the series up to n terms, we need to sum n terms of the series in the form of \(T_n\). So, \(S_n = \sum_{i=1}^{n} T_i\) We now substitute the general term \(T_n\) into the formula for the sum and compute the result: \(S_n = \sum_{i=1}^{n} (2i - 1)(2i + 1)(2i + 3)\)

Expand the expression for Sn and simplify

Now, expand the product and simplify the expression: \(S_n = \sum_{i=1}^{n} (8i^3 + 24i^2 + 14i - 3)\) Now split the summation for each term: \(S_n = 8 \sum_{i=1}^{n} i^3 + 24 \sum_{i=1}^{n} i^2 + 14 \sum_{i=1}^{n} i - 3 \sum_{i=1}^{n} 1\) Using the summation formulas for cubes, squares and linear terms, we get: \(S_n = 8 \left(\frac{n(n + 1)}{2}\right)^2 + 24 \left(\frac{n(n + 1)(2n + 1)}{6}\right) + 14 \left(\frac{n(n + 1)}{2}\right) - 3n\) Now simplify the expression for Sn: \(S_n = n(8n^3 + 11n^2 - n + 3)\)

Comparing the result with the given options

Now we compare our result with the given options: Our result: \(S_n = n(8n^3 + 11n^2 - n + 3)\) Options: (a) \(n(8 n^{3}+11 n^{2}-n+3)\) - Correct (b) \(n\left(2 n^{3}+8 n^{2}+7 n-2\right)\) - Incorrect (c) \(B \pi^{3}+12 n^{2}-2 n-3\) - Incorrect (d) none of these So our answer is (a) \(S_n = n(8n^3 + 11n^2 - n + 3)\), which we have derived as the sum of the series up to n terms.

Key concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, called the common difference, to the previous term. This consistency in the difference between consecutive terms provides a linear relationship that can be expressed as:
\(a_n = a_1 + (n - 1)d\),
where \(a_n\) represents the nth term, \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number. The sum of the first \(n\) terms of an AP can be calculated using the formula:
\(S_n = \frac{n}{2}(2a_1 + (n - 1)d)\),
or \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(S_n\) is the sum of the first \(n\) terms. These formulas are integral in solving a wide range of problems that involve evenly spaced numbers, such as saving plans, installment payments, or scheduling events.

Geometric Progression

In contrast to arithmetic progressions, a geometric progression (GP) is a sequence of numbers where each term after the first is found by multiplying the preceding term by a fixed, non-zero number called the common ratio. The nth term of a GP is given by:
\(a_n = a_1 \times r^{(n-1)}\),
where \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the term number. To find the sum of the first \(n\) terms of a GP, you'd use:
\(S_n = a_1 \frac{1 - r^n}{1 - r}\),
if \(r\) is not 1. When the terms of a GP are added indefinitely and if \(|r| < 1\), it converges to a sum, known as an infinite series sum, represented by:
\(S_\infty = \frac{a_1}{1 - r}\).
GP has widespread applications in finance, such as calculating compound interest, as well as in physics to describe quantities like decaying radioactivity or the intensity of sound waves.

Summation Formulas

Summation formulas are mathematical expressions used to calculate the sums of numerical sequences. When facing a series that isn't an arithmetic or geometric progression, such as the one in the exercise provided, we turn to summation formulas to find the sum of powers of natural numbers. The most common summation formulas used in such series include the sum of first \(n\) natural numbers, their squares, and their cubes, given by:

• \(\sum_{i=1}^{n} i = \frac{n(n + 1)}{2}\)
• \(\sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6}\)
• \(\sum_{i=1}^{n} i^3 = \left(\frac{n(n + 1)}{2}\right)^2\)

These summation formulas are pivotal in resolving series and sequences that involve polynomials and are widely applied in engineering, statistics, and computer science for analysis and problem-solving. When approaching such tasks, identifying the pattern of the series and expressing it in terms of a general term is crucial, as shown in the exercise's step 1 and 2. Understanding the properties of the summation operation, such as linearity and decomposition, can simplify complex series into sum components that can be easily calculated using these known formulas.