The product 2^(1/4) * 4^(1/16) * 8^(1/48) ... to ∞ is equal to: (A) 2^(1/4) (B) 2 (C) 2^(1/2) (D) 1

Q: Wait, why can't I just multiply the bases $2 \times 4 \times 8$ and then add the exponents?

That is a common trap! Multiplication of bases is only valid if the exponents are the same, or if you are multiplying the entire terms. Here, the exponents are different. We must use the property $a^m \times a^n = a^{m+n}$ after converting everything to the same base.

Q: How do I know the infinite series doesn't just go to infinity?

The sum of a geometric series converges if the absolute value of the common ratio $|r| < 1$. In our case, $r = \frac{1}{2}$. Since $| \frac{1}{2} | < 1$, the series is guaranteed to converge to a finite value.

Q: Is there a shortcut to identifying the GP?

Always look at the ratio of consecutive terms. If the ratio is constant, it is a GP. Here, $\frac{1/8}{1/4} = \frac{1}{2}$ and $\frac{1/16}{1/8} = \frac{1}{2}$. Since the ratio is constant, it is definitely a GP.

Question

The product $2^{1/4} 	imes 4^{1/16} 	imes 8^{1/48} \dots$ to $\infty$ is equal to:

Options:

(A) $2^{1/4}$

(B) $2$

(C) $2^{1/2}$

(D) $1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the edge of an infinite abyss, looking at a sequence of numbers that seem to grow, yet somehow, they are bound by a hidden, elegant order. We are presented with the product:

$$P = 2^{\frac{1}{4}} 	imes 4^{\frac{1}{16}} 	imes 8^{\frac{1}{48}} 	imes \dots \infty$$

At first glance, it looks like a chaotic mess of powers. But in the world of JEE mathematics, chaos is just order waiting to be discovered. Our mission is to tame this infinite product.

### Unmasking the Bases

The first step in any complex problem is to find a common language. Here, the bases are $2, 4, 8, \dots$. If you look closely, you will see they are all powers of $2$.

We know that $4 = 2^2$ and $8 = 2^3$. By rewriting the expression, we transform the problem into a uniform landscape:

$$P = 2^{\frac{1}{4}} 	imes (2^2)^{\frac{1}{16}} 	imes (2^3)^{\frac{1}{48}} 	imes \dots \infty$$

This is the moment the problem begins to yield. We have moved from a collection of different bases to a single, powerful base of $2$.

### The Exponent Dance

Now, we apply the fundamental law of exponents: $(a^m)^n = a^{mn}$. This rule allows us to simplify the powers.

For the second term, we have:

$$(2^2)^{\frac{1}{16}} = 2^{\frac{2}{16}} = 2^{\frac{1}{8}}$$

For the third term, we have:

$$(2^3)^{\frac{1}{48}} = 2^{\frac{3}{48}} = 2^{\frac{1}{16}}$$

Suddenly, the exponents are not just random fractions; they are $\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$. The pattern is beautiful, isn't it? We are now looking at:

$$P = 2^{\frac{1}{4}} 	imes 2^{\frac{1}{8}} 	imes 2^{\frac{1}{16}} 	imes \dots \infty$$

When we multiply terms with the same base, we add the exponents:

$$P = 2^{\left( \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots \infty ight)}$$

### The Infinite Convergence

We have reduced the entire infinite product to a single exponent: the sum of an infinite geometric progression (G.P.). The first term $a$ is $\frac{1}{4}$, and the common ratio $r$ is $\frac{1}{2}$.

Because $|r| < 1$, we can use the elegant formula for the sum of an infinite G.P.:

$$S_{\infty} = \frac{a}{1-r}$$

Substituting our values, we get:

$$S_{\infty} = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$$

The infinite sum collapses into a simple fraction, $\frac{1}{2}$.

### The Final Revelation

We return to our base. The entire product $P$ is simply $2$ raised to the power of our sum, $S_{\infty}$.

Thus, $P = 2^{\frac{1}{2}}$, which is the square root of $2$. We started with an infinite, intimidating expression and ended with a simple, elegant constant.

The final result is:

$$\mathbf{P = \sqrt{2}}$$

This is the beauty of mathematics—taking the infinite and making it finite, taking the complex and making it simple. You have successfully navigated the trap and found the truth.

Animated Solution for Mathematics - Sequence and Series: The product $2^{1/4} \times 4^{1/16} \times 8^{1/48} \dots$ to $\infty$ is equal to:

Select Answer:

Visualized Solution

Analyze the Infinite Product

Express Bases as Powers of $2$

Simplify the Exponents

Combine Powers with Same Base

Identify the Infinite G.P.

Visualize the First Term

Visualize the Subsequent Terms

Sum of Infinite G.P. Formula

Calculate the Exponent Sum

Final Result

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

Unmasking the Bases

The Exponent Dance

The Infinite Convergence

The Final Revelation

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