If the value of (1 + 2/3 + 6/3² + 10/3³ + ... upto ∞)^log(0.25)(1/3 + 1/3² + 1/3³ + ... upto ∞) is l

Q: Wait, why do we separate the '1' from the rest of the series in the base?

That is a brilliant observation! If you look closely at the series $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \dots$, the numerators are $1, 2, 6, 10, \dots$. This sequence is not a pure AP because of that initial $1$. By isolating it, we reveal the underlying structure: $2, 6, 10, \dots$, which is a perfect AP with common difference $4$. It makes the math much cleaner and prevents errors in the shift-and-subtract step.

Q: Why does the base of the logarithm $0.25$ become $1/2$ in the final exponent calculation?

Great question! We use the property $\log_{a^k} b = \frac{1}{k} \log_a b$. Here, the base is $0.25$, which is $\frac{1}{4}$. We can write $\frac{1}{4}$ as $(\frac{1}{2})^2$. So, $\log_{(1/2)^2} (\frac{1}{2})$ becomes $\frac{1}{2} \log_{1/2} (\frac{1}{2})$. Since $\log_x x = 1$, the whole expression simplifies beautifully to $\frac{1}{2}$.

Q: Is there a shortcut formula for the sum of an infinite AGP?

Yes, there is a general formula, but I strongly advise against memorizing it for JEE Advanced. The 'Shift-and-Subtract' method is a fundamental technique that works even when the series is slightly modified. It builds your mathematical intuition and is much safer than relying on a formula you might misremember under exam pressure. Stick to the method, and you will never go wrong!

Question

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

Options:

(A) 6

(B) 4

(C) 3

(D) 8

Sigma Percentile · Accepted Answer

### The Art of Deconstruction: Taming the Mathematical Monster

Have you ever stared at a problem that looked like a tangled knot of numbers, logs, and infinite series, and felt that familiar sinking feeling? You are not alone.

In JEE Advanced, examiners love to present problems that look like monsters, but beneath the surface, they are often just elegant puzzles waiting to be solved. Today, we are going to deconstruct one such problem. We will peel back the layers, simplify the chaos, and find the beauty in the calculation.

### Phase 1: The Base (The AGP)

Our expression is $l = S^E$. Let us first focus on the base, $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \dots \infty$.

As we noted, that initial $1$ is a bit of a distractor. Let us set it aside and focus on $S' = \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \dots \infty$.

Look at the numerators: $2, 6, 10, \dots$. This is an arithmetic progression with a first term $a=2$ and a common difference $d=4$.

Look at the denominators: $3, 3^2, 3^3, \dots$. This is a geometric progression with a common ratio $r=\frac{1}{3}$. When you have an AP multiplied by a GP, you have an Arithmetico-Geometric Progression (AGP).

The standard, most reliable way to solve this is the 'Shift-and-Subtract' method. We write $S'$ and then write $\frac{1}{3}S'$ (the common ratio multiplied by the series), shifting the terms one position to the right:

$$S' = \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \dots$$

$$\frac{1}{3}S' = \frac{2}{3^2} + \frac{6}{3^3} + \dots$$

Now, subtract the second from the first. On the left, we have $S' - \frac{1}{3}S' = \frac{2}{3}S'$.

On the right, the first term $\frac{2}{3}$ stays, and the subsequent terms subtract beautifully: $\frac{6-2}{3^2} + \frac{10-6}{3^3} + \dots$, which simplifies to $\frac{4}{3^2} + \frac{4}{3^3} + \dots$. We have successfully reduced the AGP to a simple infinite GP!

Using the sum formula $S_{\infty} = \frac{a}{1-r}$, where $a = \frac{4}{9}$ and $r = \frac{1}{3}$, we get:

$$\frac{4/9}{1-1/3} = \frac{4/9}{2/3} = \frac{2}{3}$$

Thus, $\frac{2}{3}S' = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$, which means $S' = 2$. Adding back our initial $1$, we find the base $S = 1 + 2 = 3$.

### Phase 2: The Exponent (The Logarithmic Twist)

Now, let us tackle the exponent $E = \log_{0.25} (\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots \infty)$.

The argument inside the log is a straightforward infinite GP with $a = \frac{1}{3}$ and $r = \frac{1}{3}$. Its sum is:

$$\frac{1/3}{1-1/3} = \frac{1/3}{2/3} = \frac{1}{2}$$

Now we have $E = \log_{0.25} (\frac{1}{2})$. Remember that $0.25 = \frac{1}{4} = (\frac{1}{2})^2$.

Using the logarithmic property $\log_{a^k} b = \frac{1}{k} \log_a b$, we get $E = \frac{1}{2} \log_{1/2} (\frac{1}{2})$. Since $\log_{1/2} (\frac{1}{2}) = 1$, the exponent simplifies to $E = \frac{1}{2}$.

### Phase 3: The Synthesis

We have arrived at the final stage. We found $S = 3$ and $E = \frac{1}{2}$.

The expression $l$ is simply $3^{1/2}$, or $\sqrt{3}$. The question asks for $l^2$.

Squaring $\sqrt{3}$ gives us **3**.

Animated Solution for Mathematics - Sequence and Series: 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

Select Answer:

Visualized Solution

Structure of the Expression

Identifying the Base Sequence

Setting up the AGP Calculation

Applying the Shift-and-Subtract Method

Summing the Resulting GP

Finalizing the Base Value $S$

Evaluating the Exponent's Argument

Solving the Logarithmic Exponent

Combining Base and Exponent

The Way Forward

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

The Art of Deconstruction

Phase 1

Phase 2

Phase 3

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