The product of 2^(1/4) * 4^(1/16) * 8^(1/48) * 16^(1/128) * .....∞ is (A) √2 (B) 2 (C) 2^(1/2) (D) 1

Q: Wait, why can't we just multiply the bases together first? Like $(2 \cdot 4 \cdot 8 \dots)$?

That is a classic trap! Exponent laws do not allow us to multiply bases when the exponents are different. The rule $a^m \cdot b^m = (ab)^m$ only works if the exponents are identical. Here, the exponents are different, so we must force the bases to be identical first. Always seek a common base when dealing with exponential products.

Q: How do we know the infinite sum in the exponent actually converges?

Great question! An infinite Geometric Progression converges if and only if the absolute value of the common ratio $|r|$ is less than $1$. In our series, the common ratio is $r = \frac{1}{2}$. Since $|\frac{1}{2}| < 1$, the sum is guaranteed to converge to a finite value. You are safe to use the formula!

Q: Is there a shortcut to seeing that the exponents simplify to a G.P.?

The shortcut is pattern recognition. When you see terms like $\frac{1}{4}, \frac{2}{16}, \frac{3}{48}$, look at the relationship between the numerator and the denominator. Notice that the denominator is growing much faster than the numerator. By simplifying each fraction individually, the hidden G.P. structure will always emerge. Trust the algebra!

Question

The product of $2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot .....\infty$ is

Options:

(A) $\sqrt{2}$

(B) 2

(C) $2^{\frac{1}{2}}$

(D) 1

Sigma Percentile · Accepted Answer

### The Infinite Mystery

Welcome, future engineer. Today, we aren't just solving a problem; we are unraveling a mystery.

When you see an infinite product like

$$P = 2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \dots \infty$$

your first instinct might be panic. Infinity is a big word, but in the world of JEE Advanced, infinity is often just a pattern that has decided to repeat itself forever.

### Phase 1: Finding Common Ground

Look at the bases: $2, 4, 8, 16$. They look different, but they are all powers of $2$.

This is our **'Aha!' moment**. We can rewrite them as $2^1, 2^2, 2^3, 2^4$.

By substituting these back into our expression, we transform the problem from a chaotic mix of numbers into a unified family of powers. We are no longer dealing with different bases; we are dealing with the base $2$ in various disguises.

### Phase 2: The Power of Powers

Now, we apply the law $(a^m)^n = a^{m \cdot n}$. This is the engine of our solution.

When we raise $(2^2)$ to the power of $\frac{1}{16}$, we get $2^{\frac{2}{16}}$. When we raise $(2^3)$ to the power of $\frac{1}{48}$, we get $2^{\frac{3}{48}}$.

Suddenly, the exponents are no longer just fractions; they are products waiting to be simplified. Let's do the math:

$$\frac{2}{16} = \frac{1}{8}, \quad \frac{3}{48} = \frac{1}{16}, \quad \frac{4}{128} = \frac{1}{32}$$

Look at that! The pattern is emerging: $\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32} \dots$. It is a beautiful, descending sequence.

### Phase 3: The Infinite Geometric Progression

Since all our bases are now $2$, we use the product rule $a^m \cdot a^n = a^{m+n}$.

Our entire infinite product collapses into a single base of $2$ raised to the sum of these fractions:

$$S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \dots \infty$$

This is an infinite **Geometric Progression (G.P.)**. The first term $a = \frac{1}{4}$, and the common ratio $r = \frac{1}{2}$.

Because $|r| < 1$, we can use the elegant sum formula:

$$S = \frac{a}{1-r}$$

### The Final Reveal

Substituting our values, we get:

$$S = \frac{\frac{1}{4}}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$$

Our terrifying infinite exponent is just $\frac{1}{2}$.

Therefore, our final answer is $P = 2^{\frac{1}{2}}$, which is simply **$\sqrt{2}$**.

You see? What looked like an insurmountable mountain of infinity was just a well-behaved geometric series in disguise. Keep this logic in your toolkit—whenever you see infinite products, look for the common base and the hidden series.

Animated Solution for Mathematics - Sequence and Series: The product of $2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 1 6^{\frac{1}{128}} \cdot .....\infty$ is

Select Answer:

Visualized Solution

Analyze the Infinite Product

Convert Bases to Power of $2$

Apply Power Rule $(a^{m})^{n} = a^{mn}$

Simplify Fractional Exponents

Combine Powers

Identify the Infinite G.P.

Visualize the Infinite Sum

Sum of Infinite G.P. Formula

Calculate the Exponent Sum

Final Result

The Sigma Insight: Geometric Progression (G.P.)

The Infinite Mystery

Phase 1

Phase 2

Phase 3

The Final Reveal

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The product of 241​⋅4161​⋅8481​⋅161281​⋅.....∞ is