If the sum of the first ten terms of the series 1/5 + 2/65 + 3/325 + 4/1025 + 5/2501 + ... is m/n, w

Q: Wait, how was I supposed to know to add and subtract $4n^2$? That feels like a random magic trick.

That is a very fair frustration! This is known as the Sophie Germain Identity. In JEE Advanced, you will often encounter expressions of the form $a^4 + 4b^4$. The trick is to complete the square by adding and subtracting $2a^2b^2$. It is a specific tool in your mathematical toolkit—once you see it, you will recognize it instantly in future problems.

Q: Why does the telescoping series work here? I always get confused about which terms cancel out.

Think of it like a line of dominoes. When you write the sum as $S = \sum [f(n) - f(n+1)]$, the terms look like $(f(1) - f(2)) + (f(2) - f(3)) + (f(3) - f(4)) + \dots$. Notice how the $-f(2)$ and $+f(2)$ annihilate each other? The only things left standing are the very first term and the very last term. Just write out the first three terms explicitly, and the pattern will reveal itself.

Q: How do I verify if $55$ and $221$ are co-prime? Is there a quick way?

Great question! To check for co-primality, look at the prime factors. $55$ is $5 \times 11$. Now, check if $221$ is divisible by $5$ (no, it doesn't end in $0$ or $5$) or $11$ (the alternating sum of digits $2-2+1 = 1$, which is not $0$ or $11$). Since neither factor divides $221$, they are indeed co-prime. You can proceed with confidence!

Question

If the sum of the first ten terms of the series $\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \dots$ is $\frac{m}{n}$, where $m$ and $n$ are co-prime numbers, then $m + n$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Series

Have you ever looked at a series and felt like it was speaking a secret language? When you first encounter the series

$$\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \dots$$

it might look like a chaotic mess of numbers. But in the world of JEE Advanced, chaos is just order waiting to be discovered. Let us embark on a journey to decode the DNA of this series.

### Decoding the DNA

To solve any series, we must find the general term, $T_n$. The numerators are simple: $1, 2, 3, \dots$, which is just $n$.

The denominators are $5, 65, 325, \dots$. By observing the pattern, we find that $5 = 4(1)^4 + 1$, $65 = 4(2)^4 + 1$, and $325 = 4(3)^4 + 1$.

Suddenly, the pattern emerges from the fog:

$$T_n = \frac{n}{4n^4 + 1}$$

We have successfully identified the heart of the problem.

### The Sophie Germain Magic

Now, we face the beast: $4n^4 + 1$. To factorize this quartic polynomial, we use a legendary algebraic maneuver known as the **Sophie Germain Identity**.

We want to complete the square, so we cleverly add and subtract $4n^2$:

$$4n^4 + 1 + 4n^2 - 4n^2 = (2n^2 + 1)^2 - (2n)^2$$

This is the difference of two squares. Applying $a^2 - b^2 = (a-b)(a+b)$, we get:

$$(2n^2 - 2n + 1)(2n^2 + 2n + 1)$$

The beast is now tamed.

### The Partial Fraction Split

With our denominator factorized, we perform partial fraction decomposition. We have:

$$T_n = \frac{n}{(2n^2 - 2n + 1)(2n^2 + 2n + 1)}$$

Notice the difference between the two factors: $(2n^2 + 2n + 1) - (2n^2 - 2n + 1) = 4n$. Since our numerator has an $n$, we multiply and divide by $4$ to balance the expression.

This gives us:

$$T_n = \frac{1}{4} \left[ \frac{1}{2n^2 - 2n + 1} - \frac{1}{2n^2 + 2n + 1} \right]$$

This is the "Aha!" moment. We have transformed a complex fraction into a beautiful difference of two terms.

### The Telescoping Symphony

Now, let us watch the magic unfold. When we sum these terms from $n=1$ to $n=10$, we get a chain reaction.

For $n=1$, we have $\frac{1}{4} [1 - \frac{1}{5}]$. For $n=2$, we have $\frac{1}{4} [\frac{1}{5} - \frac{1}{13}]$.

The negative $\frac{1}{5}$ from the first term cancels the positive $\frac{1}{5}$ from the second. This continues like falling dominoes.

Every intermediate term vanishes, leaving only the first positive fraction and the very last negative fraction. After the dust settles, we are left with:

$$\frac{1}{4} \left[ 1 - \frac{1}{2(10)^2 + 2(10) + 1} \right] = \frac{1}{4} \left[ 1 - \frac{1}{221} \right]$$

### Final Calculation

Calculating this is the final step of our journey.

$$1 - \frac{1}{221} = \frac{220}{221}$$

Multiplying by the $\frac{1}{4}$ outside, we get:

$$\frac{55}{221}$$

Since $55$ and $221$ are co-prime, we have found our $m=55$ and $n=221$. The sum **$m+n = 276$**. You have not just solved a problem; you have mastered the art of pattern recognition.

Animated Solution for Mathematics - Sequence and Series: If the sum of the first ten terms of the series $\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \dots$ is $\frac{m}{n}$ , where $m$ and $n$ are co-prime numbers, then $m + n$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Identifying the General Term $T_{n}$

The Sophie Germain Identity

Factorizing the Denominator

Setting up Partial Fractions

Splitting the Term

Expanding the Sum for $n = 1$ and $n = 2$

Expanding the Sum up to $n = 10$

The Telescoping Magic

The Surviving Terms

Calculating the Final Sum

Finding $m + n$

The Sigma Insight: Sum of Special Series

Analyzing the Series

Decoding the DNA

The Sophie Germain Magic

The Partial Fraction Split

The Telescoping Symphony

Final Calculation

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