Question

The sum to infinity of the arithmeticogeometric series \(1+3 \times \frac{1}{2}+5 \times \frac{1}{4}+7 \times \frac{1}{8}+9 \times \frac{1}{16}+\ldots \ldots \ldots .\) to \(\infty\) is (a) \(\frac{1}{4}\) (b) \(\frac{5}{2}\) (c) 6 (d) 7

Short answer

\(1+3 \times \frac{1}{2}+5 \times \frac{1}{4}+7 \times \frac{1}{8}+9 \times \frac{1}{16}+\ldots\) Options: (a) \(\frac{3}{2}\) (b) \(\frac{5}{2}\) (c) \(\frac{4}{5}\) (d) \(\frac{5}{4}\) Answer: (d) \(\frac{5}{4}\)

Step-by-step solution

Identify arithmetic and geometric sequences

In the given series \(1+3 \times \frac{1}{2}+5 \times \frac{1}{4}+7 \times \frac{1}{8}+9 \times \frac{1}{16}+\ldots\), we can see two different sequences: - The arithmetic sequence is given by the odd numbers: \(1, 3, 5, 7, 9, ...\) - The geometric sequence is given by the powers of \(\frac{1}{2}\): \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, ...\)

Express the series as a summation

Let's express the arithmetic-geometric series as: \(S = \sum_{n=1}^{\infty} (2n-1) \times \frac{1}{2^{n}}\) Now we'll deal with the sum and their respective sequences separately.

Calculate the sum of the arithmetic sequence

The arithmetic sequence can be expressed using the general formula \(a_n = a_1 + (n-1)d\), where \(n\) is the term number, \(a_n\) is the value of the term, \(a_1\) is the first term, and \(d\) is the common difference: \(a_n = 1 + (n-1)2 = 2n - 1\)

Calculate the sum of the geometric sequence

Since it's an infinite geometric series, we can use the sum formula \(S_{\infty} = \frac{a_1}{1 - r}\), where \(a_1\) is the first term and \(r\) is the common ratio: \(S_{\infty} = \frac{1}{1 - \frac{1}{2}} = 2\)

Combine the arithmetic and geometric sums

Now we can combine the sums of the two sequences and rewrite the given series as: \(S = \sum_{n=1}^{\infty} (2n-1) \times \frac{1}{2^{n}} = \sum_{n=1}^{\infty} (2n-1) \times 2^{-n}\)

Evaluate the summation

To evaluate the summation, we can use differentiation and the geometric series sum: \(S = \sum_{n=1}^{\infty} (2n-1) \times 2^{-n} = \frac{d}{dr}\sum_{n=1}^{\infty} (2n-1)r^n\) Differentiating and applying the geometric series sum formula: \(S = \frac{d}{dr}\left(\frac{r}{(1 - r)^2} - \frac{1}{(1 - r)^2}\right)\) Evaluate the derivative and plug in \(r=\frac{1}{2}\): \(S = \frac{5}{4}\) Based on the calculation, the correct option is: (b) \(\frac{5}{2}\)

Key concepts

Arithmetic Sequence

An arithmetic sequence is a list of numbers where each term after the first is obtained by adding a constant difference to the preceding term. This constant difference is known as the "common difference." For example, in the arithmetic sequence given in the problem: 1, 3, 5, 7, 9,..., the first term is 1 and the common difference is 2.

• The general formula for the nth term of an arithmetic sequence is: \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term and \(d\) is the common difference.
• Using the formula, we can express any term in the sequence in terms of n, like \(a_n = 2n - 1\) in the given example.

This sequence grows linearly, which means it will continue infinitely in the positive direction if not bounded. In the context of the exercise, the arithmetic sequence is an integral part when combined with the geometric sequence to form an arithmetic-geometric series.

Geometric Sequence

A geometric sequence is a series of numbers where each term after the first is found by multiplying the previous one with a fixed, non-zero number called the "common ratio." In the exercise, the sequence \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16},...\) represents a geometric sequence.

• The first term \((a_1)\) is \(\frac{1}{2}\).
• The common ratio \((r)\) is also \(\frac{1}{2}\).

The general term for a geometric sequence is \(a_n = a_1 \times r^{(n-1)}\).
For any geometric sequence, if you have the first term and the common ratio, you can calculate any term in the sequence. The neat thing about geometric sequences is that they can potentially sum to a finite value even if they go on forever (sum to infinity). This characteristic is particularly useful in dealing with infinite series.

Sum to Infinity

The sum to infinity is the sum of all terms of an infinite geometric sequence. It's a powerful concept because it allows us to find a finite value for a seemingly infinite addition. For a geometric series to converge (sum to a finite value), the absolute value of the common ratio must be less than 1.

• The sum to infinity of a geometric series is calculated using the formula: \[S_{\infty} = \frac{a_1}{1 - r}\]
• Here, \(a_1\) is the first term and \(r\) is the common ratio.

In the original solution, the sum to infinity of the geometric series was calculated as 2, showing how you can obtain a finite sum from an infinite series. These kinds of sums are particularly useful in calculus and series analysis, where understanding behavior at infinity is essential.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. In this context, the exercise involves an arithmetic-geometric series, where each term is a product of the corresponding terms of an arithmetic sequence and a geometric sequence.

• The series in question can be expressed as: \[S = \sum_{n=1}^{\infty} (2n-1) \times \frac{1}{2^{n}}\]
• Handling infinite series often requires advanced techniques like differentiation and integration, along with standard formulas.

In evaluation and solutions, infinite series often appear, requiring manipulation such as differentiating and then substituting particular values, as shown in the problem's differentiation step. It opens up possibilities for calculating long-term behavior in various fields, such as physics and finance.