If the sum of the series (1/2 - 1/3) + (1/2² - 1/(2⋅3) + 1/3²) + (1/2³ - 1/(2²⋅3) + 1/(2⋅3²) - 1/3³)

Q: Why can't we just treat the whole series as one big GP?

Because the number of terms in each bracket increases as we go. It is a series of series, not a single GP. We must first find the sum of each bracket individually, and then sum those results.

Q: I keep messing up the signs in the common ratio. How do I handle $(-1/3)$?

Treat the negative sign as part of the ratio. The common ratio $r$ is the second term divided by the first term, which is $(-\frac{1}{3}) / (\frac{1}{2}) = -\frac{2}{3}$. Always keep the parentheses to avoid losing the negative sign during exponentiation.

Q: Why did the $2^n$ terms cancel out so perfectly?

That is the beauty of the problem design! It is meant to simplify. If you do not see cancellation, re-check your algebraic expansion of the finite GP sum. The terms are designed to collapse into a much simpler form.

Question

If the sum of the series $(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2}) + (\frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3}) + (\frac{1}{2^4} - \frac{1}{2^3 \cdot 3} + \frac{1}{2^2 \cdot 3^2} - \frac{1}{2 \cdot 3^3} + \frac{1}{3^4}) + \dots$ is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha + 3\beta$ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Anatomy of a Pattern

Welcome, future engineer. Today, we are going to peel back the layers of a series that, at first glance, looks like a chaotic mess of fractions. But in the world of JEE Advanced, chaos is often just order in disguise.

Let us look at the series $S = T_1 + T_2 + T_3 + \dots$. The first term is $T_1 = (\frac{1}{2} - \frac{1}{3})$. The second is $T_2 = (\frac{1}{2^2} - \frac{1}{2 \cdot 3} + \frac{1}{3^2})$. The third is $T_3 = (\frac{1}{2^3} - \frac{1}{2^2 \cdot 3} + \frac{1}{2 \cdot 3^2} - \frac{1}{3^3})$.

Do you see it? The powers of two are decreasing while the powers of three are increasing. This is not random; it is a systematic, beautiful progression.

### The GP Engine

To conquer this, we must generalize. Let us find the $n$-th term, $T_n$. If you observe the pattern, each $T_n$ is a finite geometric progression.

We can express it using summation notation:
$$T_n = \sum_{k=0}^{n} \left( \frac{1}{2} \right)^{n-k} \left( -\frac{1}{3} \right)^k$$

This is a finite GP with $n+1$ terms. The first term is $a = (\frac{1}{2})^n$, and the common ratio is $r = \frac{-1/3}{1/2} = -\frac{2}{3}$.

Now, we deploy our most trusted tool: the sum formula for a finite GP, $S_m = a \frac{1-r^m}{1-r}$, where $m = n+1$. Substituting our values, we get:
$$T_n = \left( \frac{1}{2} \right)^n \frac{1 - (-\frac{2}{3})^{n+1}}{1 - (-\frac{2}{3})}$$

The denominator simplifies to $1 + \frac{2}{3} = \frac{5}{3}$.

### The Art of Simplification

This is where the magic happens, and where you must be vigilant. We have:
$$T_n = \left( \frac{1}{2} \right)^n \frac{1 - (-\frac{2}{3})^{n+1}}{5/3}$$

Bringing the $\frac{5}{3}$ up as $\frac{3}{5}$, we get:
$$T_n = \frac{3}{5} \left[ \left( \frac{1}{2} \right)^n - \left( \frac{1}{2} \right)^n \frac{(-2)^{n+1}}{3^{n+1}} \right]$$

Watch closely as the $2^n$ terms cancel out. After careful algebraic manipulation, we arrive at the elegant form:
$$T_n = \frac{6}{5} \left[ \left( \frac{1}{2} \right)^{n+1} - \left( -\frac{1}{3} \right)^{n+1} \right]$$

This is the DNA of our series. It is no longer a monster; it is a sum of two simple, infinite geometric series.

### The Infinite Horizon

Now, we sum $T_n$ from $n=1$ to $\infty$. We can pull the constant $\frac{6}{5}$ outside:
$$S = \frac{6}{5} \left[ \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^{n+1} - \sum_{n=1}^{\infty} \left( -\frac{1}{3} \right)^{n+1} \right]$$

Let us call these $S_1$ and $S_2$. For $S_1$, the first term is $a = (\frac{1}{2})^2 = \frac{1}{4}$ and $r = \frac{1}{2}$. The sum is:
$$S_1 = \frac{1/4}{1 - 1/2} = \frac{1}{2}$$

For $S_2$, the first term is $a = (-\frac{1}{3})^2 = \frac{1}{9}$ and $r = -\frac{1}{3}$. The sum is:
$$S_2 = \frac{1/9}{1 - (-1/3)} = \frac{1/9}{4/3} = \frac{1}{12}$$

### The Victory Lap

Finally, we substitute these back:
$$S = \frac{6}{5} \left[ \frac{1}{2} - \frac{1}{12} \right]$$

Taking the common denominator, we get $\frac{6}{5} [ \frac{5}{12} ]$, which simplifies to exactly $\frac{1}{2}$. We are given $S = \frac{\alpha}{\beta} = \frac{1}{2}$, so $\alpha = 1$ and $\beta = 2$.

The final calculation, **$\alpha + 3\beta = 1 + 3(2) = 7$**, is our reward. You have just tamed a complex series using nothing but the fundamental laws of sequences. Keep this clarity, and you will solve any problem the exam throws at you.

Enter Numerical Value:

Visualized Solution

Identifying the Pattern

General Term as a GP

Applying the GP Sum Formula

Simplifying the General Term

Setting up the Infinite Sum

Splitting the Summation

Summing the First GP

Summing the Second GP

Final Series Sum

Calculating the Final Value

The Sigma Insight: Sum of Special Series

Analyzing the Anatomy of a Pattern

The GP Engine

The Art of Simplification

The Infinite Horizon

The Victory Lap

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