Key Concepts and Formulas

• Properties of Exponents:
  • $(a^m)^n = a^{mn}$
  • $a^m \cdot a^n \cdot a^p \cdot ... = a^{m+n+p+...}$
• Sum of an Infinite Geometric Progression (GP): If the first term is $a$ and the common ratio is $r$ with $|r| < 1$, then the sum to infinity is $S_\infty = \frac{a}{1-r}$.

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Step-by-Step Solution

Let the given infinite product be $P$. $P = {2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}... \text{ to } \infty$

Step 1: Express all terms with a common base. To simplify the product, we rewrite each term with the smallest common base, which is 2. We use the property $(a^m)^n = a^{mn}$.

• The first term is $2^{1/4}$.
• The second term is $4^{1/16} = (2^2)^{1/16} = 2^{2 \times (1/16)} = 2^{2/16}$.
• The third term is $8^{1/48} = (2^3)^{1/48} = 2^{3 \times (1/48)} = 2^{3/48}$.
• The fourth term is $16^{1/128} = (2^4)^{1/128} = 2^{4 \times (1/128)} = 2^{4/128}$.

Substituting these back into the product $P$: $P = {2^{{1 \over 4}}} \cdot {2^{{2 \over {16}}}} \cdot {2^{{3 \over {48}}}} \cdot {2^{{4 \over {128}}}}... \text{ to } \infty$

Step 2: Combine terms using the property of exponents. Since all terms have the same base (2), we can combine them by adding their exponents: $a^m \cdot a^n \cdot a^p \cdot ... = a^{m+n+p+...}$. $P = 2^{\left( {1 \over 4} + {2 \over {16}} + {3 \over {48}} + {4 \over {128}} + ...\infty \right)}$

Step 3: Simplify the terms in the exponent. Let the sum of the exponents be $S$. We simplify each term in the series:

• First term: $\frac{1}{4}$
• Second term: $\frac{2}{16} = \frac{1}{8}$
• Third term: $\frac{3}{48} = \frac{1}{16}$
• Fourth term: $\frac{4}{128} = \frac{1}{32}$

So, the series for the exponent becomes: $S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + {1 \over {32}} + ...\infty$

Step 4: Identify and sum the infinite geometric progression. The series $S$ is an infinite geometric progression. The first term is $a = \frac{1}{4}$. The common ratio $r$ is found by dividing any term by its preceding term: $r = \frac{{1/8}}{{1/4}} = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2}$. We verify this with the next pair of terms: $r = \frac{{1/16}}{{1/8}} = \frac{1}{16} \times 8 = \frac{8}{16} = \frac{1}{2}$. Since $|r| = |\frac{1}{2}| = \frac{1}{2} < 1$, the sum of this infinite GP converges. The sum of an infinite GP is given by $S_\infty = \frac{a}{1-r}$. Substituting the values $a = \frac{1}{4}$ and $r = \frac{1}{2}$: $S = \frac{{1 \over 4}}{1 - {1 \over 2}} = \frac{{1 \over 4}}{{1 \over 2}}$ To simplify the complex fraction, multiply the numerator by the reciprocal of the denominator: $S = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}$

Step 5: Calculate the final product. Now, substitute the sum of the exponents $S = \frac{1}{2}$ back into the expression for $P$: $P = 2^S = 2^{{1 \over 2}}$ This can also be written as $\sqrt{2}$.

Comparing this result with the given options: (A) $2^{1/4}$ (B) $2^{1/2}$ (C) 1 (D) 2

The calculated value $2^{1/2}$ matches option (B).

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Common Mistakes & Tips:

• Incorrectly Identifying the Series: After standardizing the base, carefully simplify each exponent term before attempting to identify the series. Errors in simplification can lead to misidentifying the type of series or its parameters.
• Forgetting the Conditions for Infinite GP Sum: Ensure that the absolute value of the common ratio $|r|$ is less than 1 before applying the formula $S_\infty = \frac{a}{1-r}$. If $|r| \ge 1$, the sum does not converge to a finite value.
• Algebraic Errors in Exponent Simplification: Be meticulous when simplifying fractions and performing arithmetic operations with exponents, as small errors can propagate through the calculation.

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Summary

The problem involves simplifying an infinite product by expressing all terms with a common base, 2. This transforms the product into a single power of 2, where the exponent is the sum of an infinite series. By simplifying the terms of this series, we recognize it as an infinite geometric progression with a first term of $\frac{1}{4}$ and a common ratio of $\frac{1}{2}$. Using the formula for the sum of an infinite geometric progression, we find the sum of the exponents to be $\frac{1}{2}$. Therefore, the value of the infinite product is $2^{1/2}$.

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Final Answer

The final answer is $\boxed{2^{{1 \over 2}}}$ which corresponds to option (B).