Question

Variance of \(1,3,5,7 \ldots \ldots \ldots(4 n+1)\) is (a) \([\\{2 n(2 n-1)\\} / 3]\) (c) \((1 / n) \sqrt\left[\left(n^{2}-1\right) / 3\right] 100\) (d) \([\\{4 n(n+1)\\} / 3]\)

Short answer

The variance of the sequence \(1, 3, 5, 7, \ldots, (4n + 1)\) is given by the formula \(\sigma^{2} = \frac{1}{n} \sum_{i=0}^{n-1} [(4i+1) - \frac{1}{n} \left(4 \cdot \frac{(n-1)(n)}{2} + n \right)]^{2}\).

Step-by-step solution

Find the mean, \(\bar{x}\)

To find the mean of the sequence, we first need to find the sum of the elements in the sequence divided by the number of elements in the sequence. The general form of the sequence is \(a_i = 4i + 1\), where \(i\) ranges from \(0\) to \(n - 1\). So, the sum of the sequence is: \[\sum_{i=0}^{n-1} a_i = \sum_{i=0}^{n-1} (4i + 1)\] \[= \sum_{i=0}^{n-1} 4i + \sum_{i=0}^{n-1} 1\] The sum of the sequence becomes: \[= 4 \sum_{i=0}^{n-1} i + n\] By using the formula for the sum of first n natural numbers, we have \(\sum_{i=0}^{n-1} i = \frac{(n-1)(n)}{2}\). The mean, \(\bar{x}\), is: \[\bar{x} = \frac{1}{n} \left(4 \cdot \frac{(n-1)(n)}{2} + n \right)\]

Calculate the sum of squared differences, \(\sum_{i=1}^{n} (x_i - \bar{x})^{2}\)

Now, let's calculate the sum of squared differences: \[\sum_{i=0}^{n-1} (a_i - \bar{x})^{2} = \sum_{i=0}^{n-1} [(4i+1) - \frac{1}{n} \left(4 \cdot \frac{(n-1)(n)}{2} + n \right)]^{2}\]

Find the variance using the formula, \(\sigma^{2}\)

Finally, we can find the variance of the sequence using the formula: \[\sigma^{2} = \frac{1}{n} \sum_{i=0}^{n-1} [(4i+1) - \frac{1}{n} \left(4 \cdot \frac{(n-1)(n)}{2} + n \right)]^{2}\] By calculating the variance of the given sequence with the provided formula, we can compare our results to the given options (a), (c), and (d).

Key concepts

Arithmetic Sequence

An arithmetic sequence is a series of numbers in which each term after the first is obtained by adding a constant difference to the preceding term. This difference is known as the common difference, denoted typically by 'd'. For instance, in the sequence 1, 3, 5, 7, the common difference is 2, because each term is increased by 2 to get the next term.

An arithmetic sequence can be expressed in the form of a general term, denoted as \( a_n = a_1 + (n - 1)d \), where \( a_n \) is the nth term of the sequence, \( a_1 \) is the first term, and 'n' is the number of terms. In our given exercise, the sequence is generated by \( a_i = 4i + 1 \), showcasing that we have an arithmetic sequence with a common difference of 4.

Mean of a Sequence

The mean of a sequence, also known as the average, is the sum of all the terms divided by the number of terms. For an arithmetic sequence, calculating the mean involves adding together all the terms and dividing by the total count of those terms.

In mathematical terms, the mean \( \bar{x} \) is represented as \( \bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i \), where 'n' is the number of terms and \( x_i \) represents each term of the sequence. The mean gives us a central value of the sequence and is used extensively in statistical analysis, as seen in the variance calculation of our exercise.

Sum of Natural Numbers

The sum of the first 'n' natural numbers is the total when adding all the whole numbers from 1 to 'n'. This sum can be found quickly using the formula \( S_n = \frac{n(n+1)}{2} \), which is derived from the pairing method initially conceived by Carl Friedrich Gauss, a renowned mathematician.

For example, to find the sum of the first 5 natural numbers, you would use the formula like so: \( S_5 = \frac{5(5+1)}{2} = \frac{5\times6}{2} = 15 \). This formula is especially useful when dealing with large sequences, such as in our variance calculation, where we sum a sequence of terms in an arithmetic progression.