The value of the infinite product \({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \) is
36
Calculating the Value of an Infinite Product
The problem asks for the value of an infinite product involving the base 6 raised to a sequence of exponents.
The given infinite product is:
\({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \)
Using the property of exponents, \(a^m \times a^n = a^{m+n}\), an infinite product of the form \(\prod_{n=1}^{\infty} a^{x_n}\) can be written as \(a^{\sum_{n=1}^{\infty} x_n}\). In this case, the base is \(a=6\), and the exponents are \(x_n\). So, the value of the product is \(6^S\), where \(S\) is the sum of the infinite series of exponents.
The sequence of exponents is:
- \(a_1 = \frac{1}{2}\)
- \(a_2 = \frac{1}{2}\)
- \(a_3 = \frac{3}{8}\)
- \(a_4 = \frac{1}{4}\)
- and so on...
We need to find the sum \(S = \frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \ldots\).
Finding the Pattern in the Exponents Sequence
Let's examine the given terms in the sequence of exponents to find a pattern. Let the sequence be denoted by \(a_n\).
\(a_1 = \frac{1}{2}\)
\(a_2 = \frac{1}{2}\)
\(a_3 = \frac{3}{8}\)
\(a_4 = \frac{1}{4}\)
Let's try to find if this sequence follows a linear recurrence relation of the form \(a_n = c_1 a_{n-1} + c_2 a_{n-2}\) for \(n \ge 3\). We can use the given terms to set up a system of equations for the coefficients \(c_1\) and \(c_2\).
- Using \(a_1, a_2, a_3\): For \(n=3\), \(a_3 = c_1 a_2 + c_2 a_1\). Substituting the values, we get:
\(\frac{3}{8} = c_1 \left(\frac{1}{2}\right) + c_2 \left(\frac{1}{2}\right)\)
Multiplying the equation by 8 to clear the denominators:
\(3 = 4c_1 + 4c_2\)
Dividing by 4:
\(c_1 + c_2 = \frac{3}{4}\) (Equation 1) - Using \(a_2, a_3, a_4\): For \(n=4\), \(a_4 = c_1 a_3 + c_2 a_2\). Substituting the values, we get:
\(\frac{1}{4} = c_1 \left(\frac{3}{8}\right) + c_2 \left(\frac{1}{2}\right)\)
Multiplying the equation by 8 to clear the denominators:
\(2 = 3c_1 + 4c_2\) (Equation 2)
Solving for the Recurrence Coefficients
Now we solve the system of linear equations for \(c_1\) and \(c_2\):
\(c_1 + c_2 = \frac{3}{4}\)
\(3c_1 + 4c_2 = 2\)
From Equation 1, we can express \(c_1\) as \(c_1 = \frac{3}{4} - c_2\). Substitute this into Equation 2:
\(3\left(\frac{3}{4} - c_2\right) + 4c_2 = 2\)
\(\frac{9}{4} - 3c_2 + 4c_2 = 2\)
\(c_2 = 2 - \frac{9}{4} = \frac{8}{4} - \frac{9}{4} = -\frac{1}{4}\)
Now substitute the value of \(c_2\) back into the expression for \(c_1\):
\(c_1 = \frac{3}{4} - c_2 = \frac{3}{4} - \left(-\frac{1}{4}\right) = \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1\)
So, the recurrence relation for the exponents is \(a_n = 1 \cdot a_{n-1} - \frac{1}{4} a_{n-2}\), which simplifies to \(a_n = a_{n-1} - \frac{1}{4} a_{n-2}\) for \(n \ge 3\). The initial conditions are \(a_1 = \frac{1}{2}\) and \(a_2 = \frac{1}{2}\).
Summing the Infinite Series of Exponents
We need to find the sum \(S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \ldots\). The terms \(a_n\) are generated by the recurrence \(a_n - a_{n-1} + \frac{1}{4} a_{n-2} = 0\) for \(n \ge 3\).
Assuming the series converges, we can sum the recurrence relation from \(n=3\) to infinity:
\(\sum_{n=3}^{\infty} \left(a_n - a_{n-1} + \frac{1}{4} a_{n-2}\right) = \sum_{n=3}^{\infty} 0 = 0\)
This can be split into individual sums:
\(\sum_{n=3}^{\infty} a_n - \sum_{n=3}^{\infty} a_{n-1} + \frac{1}{4} \sum_{n=3}^{\infty} a_{n-2} = 0\)
Let \(S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + a_4 + \ldots\). We can express the sums in terms of \(S\) and the initial terms:
- \(\sum_{n=3}^{\infty} a_n = a_3 + a_4 + a_5 + \ldots = S - a_1 - a_2\)
- \(\sum_{n=3}^{\infty} a_{n-1} = a_2 + a_3 + a_4 + \ldots = S - a_1\)
- \(\sum_{n=3}^{\infty} a_{n-2} = a_1 + a_2 + a_3 + \ldots = S\)
Substitute these expressions back into the summed recurrence relation:
\((S - a_1 - a_2) - (S - a_1) + \frac{1}{4} S = 0\)
\(S - a_1 - a_2 - S + a_1 + \frac{1}{4} S = 0\)
The \(S\) terms cancel out, and the \(a_1\) terms cancel out:
\(-a_2 + \frac{1}{4} S = 0\)
Solving for \(S\):
\(\frac{1}{4} S = a_2\)
\(S = 4 a_2\)
Using the value of the second term, \(a_2 = \frac{1}{2}\), we find the sum of the exponents:
\(S = 4 \times \frac{1}{2} = 2\)
To confirm the series converges, we look at the characteristic equation of the recurrence: \(r^2 - c_1 r - c_2 = 0\), which is \(r^2 - 1 r - (-\frac{1}{4}) = 0\), or \(r^2 - r + \frac{1}{4} = 0\). This equation factors as \(\left(r - \frac{1}{2}\right)^2 = 0\). The roots are \(r = \frac{1}{2}\) with multiplicity 2. Since the modulus of the root \(|\frac{1}{2}| = \frac{1}{2}\) is less than 1, the terms \(a_n\) approach 0 as \(n \to \infty\), and the sum \(S\) converges to a finite value.
Calculating the Infinite Product Value
The value of the infinite product is \(6^S\), where \(S=2\).
\(6^S = 6^2 = 36\)
Final Answer
The value of the infinite product \({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \) is 36.
Revision Table: Infinite Product Calculation
| Concept | Explanation |
|---|---|
| Infinite Product Property | An infinite product of the form \(\prod_{n=1}^\infty a^{x_n}\) is equal to \(a^{\sum_{n=1}^\infty x_n}\), provided the sum of exponents converges. |
| Sum of Exponents Series | The problem reduces to finding the sum of the infinite series of exponents: \(\frac{1}{2} + \frac{1}{2} + \frac{3}{8} + \frac{1}{4} + \ldots\). |
| Linear Recurrence Relation | The sequence of exponents \((a_n)\) follows \(a_n = a_{n-1} - \frac{1}{4} a_{n-2}\) for \(n \ge 3\), with \(a_1 = \frac{1}{2}, a_2 = \frac{1}{2}\). |
| Summing Recurrence Series | For a convergent series from a homogeneous linear recurrence, summing the recurrence equation term by term can determine the total sum \(S\). In this case, \(-a_2 + \frac{1}{4} S = 0\). |
| Calculated Sum of Exponents | From \(-a_2 + \frac{1}{4} S = 0\) and \(a_2 = \frac{1}{2}\), we get \(S = 4a_2 = 4 \times \frac{1}{2} = 2\). |
| Final Product Value | The infinite product is \(6^S = 6^2 = 36\). |
Additional Information: Infinite Series and Recurrences
Infinite products are extensions of finite products. An infinite product \(\prod_{n=1}^{\infty} c_n\) converges to a non-zero value if and only if the series \(\sum_{n=1}^{\infty} (c_n - 1)\) converges. For positive terms \(c_n\), it converges if and only if \(\sum_{n=1}^{\infty} \ln(c_n)\) converges. In our case, \(c_n = 6^{a_n}\), so \(\ln(c_n) = a_n \ln(6)\). The sum \(\sum a_n \ln(6) = \ln(6) \sum a_n\). Since \(\ln(6)\) is a constant, the convergence of the infinite product depends directly on the convergence of the series of exponents \(\sum a_n\).
Linear recurrence relations describe sequences where each term is a linear combination of previous terms. The behaviour of these sequences is largely determined by the roots of their characteristic equation. For a second-order recurrence \(a_n = c_1 a_{n-1} + c_2 a_{n-2}\), the characteristic equation is \(r^2 - c_1 r - c_2 = 0\). If all roots have a magnitude less than 1, the terms \(a_n\) approach 0 as \(n\) goes to infinity, which is a necessary condition for the series \(\sum a_n\) to converge.
The method used to find the sum \(S\) by summing the recurrence relation is a valid technique when the series converges. It effectively collapses the infinite series into a simple equation involving the sum \(S\) and the initial terms of the sequence.
Similar Questions
If g is the geometric mean of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, then which one of the following is correct?
What is the greatest value of the positive integer n satisfying the condition \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}?\)
A geometric progression (GP) consists of 200 terms. If the sum of odd terms of the GP is m, and the sum of even terms of the GP is n, then what is its common ratio?
If the second term of a GP is 2 and the sum of its infinite terms is 8, then the GP is
If m is the geometric mean of \({\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)^{\log \left( {{\rm{yz}}} \right)}},{\rm{\;}}{\left( {\frac{{\rm{z}}}{{\rm{x}}}} \right)^{\log \left( {{\rm{zx}}} \right)}}{\rm{\;and\;}}{\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)^{\log \left( {{\rm{xy}}} \right)}}\) then what is the value of m?
The geometric mean of the observations x 1, x 2, x 3, … x nis G 1. The geometric mean of the observations y 1, y 2, y 3,… y nis G 2. The geometric mean of observations \(\frac{{{{\rm{x}}_1}}}{{{{\rm{y}}_1}}},\frac{{{{\rm{x}}_2}}}{{{{\rm{y}}_2}}},\frac{{{{\rm{x}}_3}}}{{{{\rm{y}}_3}}}, \ldots \frac{{{{\rm{x}}_{\rm{n}}}}}{{{{\rm{y}}_{\rm{n}}}}}\) is
If p, q, r are in one geometric progression and a, b, c are in another geometric progression, then ap, bq, cr are in
What is the sum of the series 0.5 + 0.55 + 0.555 + … to n terms?
Let t1, t2, t3 ... be in GP. What is \(\rm \left(t_1 t_3 \ldots t_{21}\right)^{\frac{1}{11}}\) equal to ?
Consider the following statements:
1. If each term of a GP is multiplied by same non-zero number, then the resulting sequence is also a GP.
2. If each term of a GP is divided by same non-zero number, then the resulting sequence is also a GP.
Which of the above statements is/are correct?
Important Questions from Geometric Progressions
If g is the geometric mean of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, then which one of the following is correct?
What is the greatest value of the positive integer n satisfying the condition \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}?\)
The sum of even numbers from 1 to 40 is:
The minimum value of the sum of real numbers a-5, a-4, 3a-3, 1, a8 and a10 with a > 0 is:
The arithmetic mean, geometric mean and median of six positive numbers a, a, b, b, c, c where a < b < c are \(\frac 7 3,\) 2, 2 respectively. Then what is the sum of the squares of all the six numbers?
