The value of 2^(1/4) . 4^(1/8) . 8^(1/16) ... inf is (A) 1 (B) 2 (C) 3/2 (D) 4

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

That is a common trap! You can only multiply bases if the exponents are the same. Here, the exponents are different for every term. We must use the law of indices $a^m \cdot a^n = a^{m+n}$ to combine them, which requires us to first make the bases identical.

Q: How do I know it is an AGP? Is there a formula I should memorize?

You don't need to memorize a complex formula for AGP. Just look at the structure: the numerators $1, 2, 3, 4$ form an Arithmetic Progression, and the denominators $4, 8, 16, 32$ form a Geometric Progression. When you see this 'product of an AP and a GP', it is an AGP. The 'shift and subtract' method is the universal tool for this.

Q: Why do we multiply by $\frac{1}{2}$? Where did that come from?

Great observation! The geometric part of our series is $\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots$. Each term is obtained by multiplying the previous one by $\frac{1}{2}$. That $\frac{1}{2}$ is the common ratio of the GP. Multiplying the entire sum by this ratio is the key to aligning the terms so they cancel out beautifully during subtraction.

Question

The value of $2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \dots \infty$ is

Options:

(A) 1

(B) 2

(C) 3/2

(D) 4

Sigma Percentile · Accepted Answer

### The Infinite Dance: Unmasking the Product

Imagine you are standing before a vast, infinite wall of numbers. You see $2^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 8^{\frac{1}{16}} \dots \infty$.

At first glance, it feels like a chaotic storm of exponents and bases. It is easy to feel small in the face of infinity, but in the world of JEE mathematics, infinity is not a wall; it is a playground. Let us break this monster down, piece by piece, until it reveals its elegant core.

### Phase 1: Finding the Common Ground

Look closely at the bases: $2, 4, 8, 16, \dots$. Do you see the hidden harmony? They are all powers of $2$.

This is our first breakthrough. We do not need to deal with different bases; we can rewrite the entire expression in the language of base $2$. We know that $4 = 2^2$, $8 = 2^3$, and so on.

Substituting these into our expression, we get:

$$P = 2^{\frac{1}{4}} \cdot (2^2)^{\frac{1}{8}} \cdot (2^3)^{\frac{1}{16}} \cdot (2^4)^{\frac{1}{32}} \dots \infty$$

Using the law of indices, $(a^m)^n = a^{m \cdot n}$, we can simplify the exponents. The expression transforms into:

$$P = 2^{\frac{1}{4}} \cdot 2^{\frac{2}{8}} \cdot 2^{\frac{3}{16}} \cdot 2^{\frac{4}{32}} \dots \infty$$

Now, we have a product of terms with the same base. When we multiply numbers with the same base, we add their exponents. Our problem has shifted from a complex product to a single base raised to an infinite sum: $P = 2^S$, where:

$$S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \dots \infty$$

### Phase 2: The AGP Technique

Now, let us focus entirely on $S$. This is the heart of the problem.

Look at the numerators: $1, 2, 3, 4, \dots$ (an Arithmetic Progression). Look at the denominators: $4, 8, 16, 32, \dots$ (a Geometric Progression with common ratio $r = \frac{1}{2}$). This is an Arithmetico-Geometric Progression (AGP).

To solve this, we use the 'shift and subtract' method. It is a beautiful, rhythmic process. First, write down the sum $S$:

$$S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \dots$$

Next, multiply the entire equation by the common ratio, $\frac{1}{2}$:

$$\frac{1}{2}S = \frac{1}{8} + \frac{2}{16} + \frac{3}{32} + \dots$$

Now, align the terms with the same denominators by shifting the second series one position to the right. Subtracting the two equations gives us:

$$S - \frac{1}{2}S = \frac{1}{4} + \left( \frac{2}{8} - \frac{1}{8} \right) + \left( \frac{3}{16} - \frac{2}{16} \right) + \left( \frac{4}{32} - \frac{3}{32} \right) + \dots$$

### Phase 3: The Grand Collapse

Look at what happens after the subtraction. The numerators all become $1$:

$$\frac{1}{2}S = \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots$$

This is no longer an AGP. It has collapsed into a simple, infinite Geometric Progression where the first term $a = \frac{1}{4}$ and the common ratio $r = \frac{1}{2}$.

The sum of an infinite GP is given by the formula $S_{\infty} = \frac{a}{1-r}$. Applying this:

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

If $\frac{1}{2}S = \frac{1}{2}$, then $S = 1$. The entire infinite sum in our exponent is just $1$.

### The Final Result

We return to our original expression $P = 2^S$. Substituting $S = 1$, we get:

$$P = 2^1 = \mathbf{2}$$

What seemed like an intimidating, infinite mountain of numbers has been reduced to a simple, elegant $2$. This is the beauty of mathematics—with the right tools and a calm mind, even the most complex problems reveal their simplicity.

Animated Solution for Mathematics - Sequence and Series: The value of $2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \dots \infty$ is

Select Answer:

Visualized Solution

The Infinite Product

Converting to a Common Base

Applying Law of Indices

Summing the Exponents

Identifying the AGP

The AGP Manipulation - Multiplication

The AGP Manipulation - Shifting and Subtracting

The Resulting Infinite GP

Sum of Infinite GP

Solving for S

Final Result

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

The Infinite Dance

Phase 1

Phase 2

Phase 3

The Final Result

Similar Questions

The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3 ^{2}} + \frac{12}{3 ^{3}} + \frac{17}{3 ^{4}} + \frac{22}{3 ^{5}} + \dots$ is equal to

The sum to infinite term of the series $1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots$ is

The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6 ^{2}} + \frac{22}{6 ^{3}} + \frac{35}{6 ^{4}} + \frac{51}{6 ^{5}} + \frac{70}{6 ^{6}} + \dots$ is equal to:

If the value of $(1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \dots upto \infty)^{l o g_{(0.25)} (\frac{1}{3} + \frac{1}{3 ^{2}} + \frac{1}{3 ^{3}} + \dots upto \infty)}$ is $l$ , then $l^{2}$ is equal to

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$ is equal to

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4 ^{2}} (3 + 2 p) + \frac{1}{4 ^{3}} (3 + 3 p) + \dots \infty$ , then the value of $p$ is_____

Let $S = 2 + \frac{6}{7} + \frac{12}{7 ^{2}} + \frac{20}{7 ^{3}} + \frac{30}{7 ^{4}} + \dots$ . then $4 S$ is equal to

Let $a_{1}, a_{2}, a_{3}, \dots$ be an AP. If $\sum_{r = 1}^{\infty} \frac{a _{r}}{2 ^{r}} = 4$ , then $4 a_{2}$ is equal to ________.

If $7 = 5 + \frac{1}{7} (5 + α) + \frac{1}{7 ^{2}} (5 + 2 α) + \frac{1}{7 ^{3}} (5 + 3 α) + \dots \infty$ , then the value of $α$ is :

Let $S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}$ . Then the value of $(16 S - (25)^{- 54})$ is equal to

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 value of 21/4⋅41/8⋅81/16…∞ is