If 1 + (√3-√2)/(2√3) + (5-2√6)/18 + (9√3-11√2)/(36√3) + (49-20√6)/180 + ... upto ∞ = 2 + (√(b/a) + 1

Q: Wait, why do we reject the negative root here? Or why did we choose $t = 1 - \sqrt{\frac{2}{3}}$?

Great question! We didn't 'reject' a root; we performed a strategic substitution. By setting $t = 1 - \sqrt{\frac{2}{3}}$, we transformed a complex, radical-heavy series into a clean, power-series form. This is a classic JEE technique: when you see a recurring pattern of radicals, define a variable to simplify the expression. It makes the subsequent calculus much cleaner.

Q: Why can't we just use energy conservation? Wait, this is a series problem, not physics. How do I know when to use partial fractions?

Haha, I love the enthusiasm! You're right, this is pure algebra. You should think about partial fractions whenever you see a product in the denominator of a summation term, like $n(n+1)$. It's a powerful tool to break a complex fraction into simpler, manageable parts that often telescope or match standard series expansions.

Q: How do I handle the $\ln(\sqrt{x})$ vs $\ln(x)$? I always get confused with log properties.

Remember the fundamental property: $\ln(x^k) = k \ln(x)$. So, $\ln(\sqrt{x})$ is just $\ln(x^{1/2})$, which becomes $\frac{1}{2} \ln(x)$. When you're matching your result to the target form, keep this property in your back pocket—it's often the final step to getting the coefficients to align perfectly.

Question

If $1 + \frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} + \frac{5-2\sqrt{6}}{18} + \frac{9\sqrt{3}-11\sqrt{2}}{36\sqrt{3}} + \frac{49-20\sqrt{6}}{180} + \dots 	ext{upto } \infty = 2 + \left(\sqrt{\frac{b}{a}} + 1ight) \log_e\left(\frac{a}{b}ight)$, where $a$ and $b$ are integers with $	ext{gcd}(a, b) = 1$, then $11a + 18b$ is equal to

Sigma Percentile · Accepted Answer

### The Infinite Series Odyssey

Welcome, warrior. Today, we stand before a mathematical beast. It is an infinite series, draped in layers of square roots and fractions, designed to make even the most seasoned student pause.

But remember, in the world of JEE Advanced, intimidation is just a test of your composure. Let us peel back the layers of this monster together.

### The Art of Substitution

Look closely at the series:

$$1 + \frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}} + \frac{5-2\sqrt{6}}{18} + \dots$$

It looks chaotic, but chaos is often just order in disguise. Let us focus on the numerator of the second term: $\sqrt{3}-\sqrt{2}$.

If we divide this by $\sqrt{3}$, we get $1 - \sqrt{\frac{2}{3}}$. Let us define this as our variable $t$. So, $t = 1 - \sqrt{\frac{2}{3}}$.

Now, look at the second term again: $\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}$. This is exactly $\frac{t}{2}$.

The pattern is emerging! If we square $t$, we get:

$$t^2 = \left(1 - \sqrt{\frac{2}{3}}\right)^2 = 1 + \frac{2}{3} - 2\sqrt{\frac{2}{3}} = \frac{5-2\sqrt{6}}{3}$$

If we divide this by $6$, we get $\frac{5-2\sqrt{6}}{18}$, which is the third term. The pattern is confirmed: the $n$-th term is $\frac{t^n}{n(n+1)}$.

### The Power of Partial Fractions

Now that we have our series $S = 1 + \sum_{n=1}^{\infty} \frac{t^n}{n(n+1)}$, we face the product in the denominator. This is where the magic of partial fractions comes in.

We know that:

$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$

This simple identity is our bridge. We can rewrite our sum as:

$$S = 1 + \sum_{n=1}^{\infty} t^n \left(\frac{1}{n} - \frac{1}{n+1}\right)$$

By distributing $t^n$, we split this into two separate sums: $\sum \frac{t^n}{n}$ and $\sum \frac{t^n}{n+1}$.

### The Logarithmic Bridge

This is the moment of truth. Do you recognize the Maclaurin series for $-\ln(1-t)$? It is $\sum_{n=1}^{\infty} \frac{t^n}{n}$.

Our first sum is exactly this! The second sum, $\sum \frac{t^n}{n+1}$, requires a little nudge.

By multiplying and dividing by $t$, we get:

$$\frac{1}{t} \sum_{n=1}^{\infty} \frac{t^{n+1}}{n+1}$$

This is almost the same log series, just missing the first term. With a bit of algebraic manipulation, we find that the entire series $S$ simplifies to:

$$S = 2 + \left(\frac{1-t}{t}\right) \ln(1-t)$$

### The Final Transformation

We are almost there. We substitute our $t$ back into the expression. With $1-t = \sqrt{\frac{2}{3}}$ and $t = 1 - \sqrt{\frac{2}{3}}$, the coefficient $\frac{1-t}{t}$ simplifies beautifully to $\sqrt{6}+2$ after rationalization.

Our series is now:

$$S = 2 + (\sqrt{6}+2) \ln\left(\sqrt{\frac{2}{3}}\right)$$

Using the log property $\ln(\sqrt{x}) = \frac{1}{2} \ln(x)$, we pull out a factor of $\frac{1}{2}$ and distribute it, giving us:

$$S = 2 + \left(\sqrt{\frac{3}{2}} + 1\right) \ln\left(\frac{2}{3}\right)$$

Comparing this to our target form $2 + (\sqrt{\frac{b}{a}} + 1) \ln(\frac{a}{b})$, we see clearly that $a=2$ and $b=3$.

The final calculation is:

$$11(2) + 18(3) = 22 + 54 = \mathbf{76}$$

We have conquered the beast. Remember, every complex problem is just a sequence of simple steps waiting for you to find the right path.

Enter Numerical Value:

Visualized Solution

Analyze the Infinite Series

Defining the Substitution $t$

Verifying the Third Term

Establishing the General Term

Partial Fraction Decomposition

Recalling the Logarithmic Series

Evaluating the Second Sum

Combining the Results

Substituting $t$ Back

Rationalizing the Coefficient

Matching the Target Form

Identifying $a$ and $b$

Final Calculation

The Sigma Insight: Sum of Special Series

The Infinite Series Odyssey

The Art of Substitution

The Power of Partial Fractions

The Logarithmic Bridge

The Final Transformation

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