If the sum of the first ten terms of the series (1 3/5)² + (2 2/5)² + (3 1/5)² + 4² + (4 4/5)² + ...

Q: Why do we have to convert to improper fractions? Can't I just square the mixed fractions as they are?

You could, but you would be making your life infinitely harder! Squaring $(1 \frac{3}{5})^2$ directly involves dealing with mixed arithmetic, which is a breeding ground for silly calculation errors. By converting to $\frac{8}{5}$, you reveal the underlying pattern of the numerators ($8, 12, 16, \dots$). In JEE Advanced, simplification is not just about making things look neat; it is about revealing the mathematical structure of the problem.

Q: Wait, why does the summation go up to $11^2$ when the question asks for the sum of the first ten terms?

That is a classic trap! Look at the general term we derived: $T_n = (\frac{4(n+1)}{5})^2$. When $n=1$, the term is $(\frac{8}{5})^2 = 2^2$. When $n=10$, the term is $(\frac{4(11)}{5})^2 = (\frac{44}{5})^2 = (\frac{44}{5})^2$ which simplifies to $11^2$. The index $n$ counts the number of terms, but the value inside the square grows with $n+1$. Always check your boundary conditions!

Q: Is there a shortcut to avoid the sum of squares formula?

While you could manually add the squares, it is inefficient and risky. The sum of squares formula is a fundamental tool in your arsenal. In competitive exams, speed and accuracy are paramount. Memorizing and correctly applying standard summation formulas is the hallmark of a prepared student. Don't look for shortcuts; look for the most robust path.

Question

If the sum of the first ten terms of the series $(1 \frac{3}{5})^2 + (2 \frac{2}{5})^2 + (3 \frac{1}{5})^2 + 4^2 + (4 \frac{4}{5})^2 + \dots$ is $\frac{16}{5} m$, then $m$ is equal to:

Options:

(A) 100

(B) 99

(C) 102

(D) 101

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The Art of Seeing Through the Noise. Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of mixed fractions.

When you see a series like $(1 \frac{3}{5})^2 + (2 \frac{2}{5})^2 + (3 \frac{1}{5})^2 + \dots$, your instinct might be to panic. But I want you to take a deep breath. In JEE Advanced, the complexity is often just a mask. Our job is to peel it away.

### The Great Simplification

Never let mixed fractions intimidate you. They are just improper fractions in disguise.

Let us convert them: $1 \frac{3}{5}$ becomes $\frac{8}{5}$, $2 \frac{2}{5}$ becomes $\frac{12}{5}$, and $3 \frac{1}{5}$ becomes $\frac{16}{5}$. Even the whole number $4$ can be written as $\frac{20}{5}$.

Suddenly, the series looks like this:

$$ \left( \frac{8}{5} ight)^2 + \left( \frac{12}{5} ight)^2 + \left( \frac{16}{5} ight)^2 + \left( \frac{20}{5} ight)^2 + \dots $$

Do you see it now? The denominator is constant, and the numerators are dancing to a rhythm of $4$. This is the beauty of pattern recognition.

### Defining the General Term

We have identified that the numerators are $8, 12, 16, 20, \dots$. These are multiples of $4$. Specifically, the $n$-th term has a numerator of $4(n+1)$.

Thus, our general term $T_n$ is:

$$ T_n = \left( \frac{4(n+1)}{5} ight)^2 $$

When we sum the first $10$ terms, we are looking at $S_{10} = \sum_{n=1}^{10} \left( \frac{4(n+1)}{5} ight)^2$. We can pull the constant $\frac{16}{25}$ out of the summation, leaving us with the sum of squares:

$$ S_{10} = \frac{16}{25} \sum_{n=1}^{10} (n+1)^2 $$

### The Boundary Trap

Here is where many students stumble. We need to sum $(n+1)^2$ for $n=1$ to $10$. This expands to $2^2 + 3^2 + \dots + 11^2$.

Our standard formula, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$, starts from $1^2$. We don't have $1^2$ in our series!

So, we calculate the sum from $1$ to $11$ and subtract $1^2$. Using the formula for $n=11$:

$$ \frac{11(12)(23)}{6} = 506 $$

Subtracting the missing $1^2$, we get $505$.

### The Final Victory

Now, we bring it all home. We have $S_{10} = \frac{16}{25} 	imes 505$.

Notice how $505$ and $25$ share a factor of $5$? Since $505 = 5 	imes 101$, canceling the $5$ leaves us with:

$$ \frac{16}{5} 	imes 101 $$

The problem states the sum is $\frac{16}{5} m$. By simple comparison, **$m = 101$**.

You have conquered the series. Keep this clarity of mind, and no problem will ever be too complex for you.

Animated Solution for Mathematics - Sequence and Series: If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

Select Answer:

Visualized Solution

Convert Mixed Fractions

Identify the Pattern

Express as a Summation

Sum of Squares Formula

Calculate Sum from $1$ to $11$

Simplify the Sum

Adjust for Starting Term

Final Computation

Solve for $m$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Great Simplification

Defining the General Term

The Boundary Trap

The Final Victory

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