If the sum of the first 10 terms of the series 4 * 1 / (1+4 * 1^4) + 4 * 2 / (1+4 * 2^4) + 4 * 3 / (1+4 * 3^4) + ..... is m/n, where gcd(m, n)=1, then m+n is equal to _______.

The correct answer is Option 441. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q22 – JEE-MAIN 2025

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Home Maths Sequences and SeriesQuestion #407

If the sum of the first 10 terms of the series

\frac{4 \cdot 1}{1 + 4 \cdot 1^{4}} + \frac{4 \cdot 2}{1 + 4 \cdot 2^{4}} + \frac{4 \cdot 3}{1 + 4 \cdot 3^{4}} + \dots

\frac{m}{n}

, where

gcd (m, n) = 1

, then

m + n

is equal to _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Series Term Structure

Analyze the general term of the series to identify a pattern that might simplify the summation.

🧠 Full Solution Path

Step 1: Identify the General Term and Factorize the Denominator✦ Active

The general term of the series can be written as $T_{k} = \frac{4 k}{1 + 4 k^{4}}$ . We can factorize the denominator using Sophie Germain's Identity, $a^{4} + 4 b^{4} = (a^{2} + 2 b^{2} + 2 a b) (a^{2} + 2 b^{2} - 2 a b)$ . For $a = 1$ and $b = k$ , we have:

1 + 4 k^{4} = (1 + 2 k^{2})^{2} - (2 k)^{2} = (1 + 2 k^{2} - 2 k) (1 + 2 k^{2} + 2 k)

So, the general term becomes:

T_{k} = \frac{4 k}{(2 k^{2} - 2 k + 1) (2 k^{2} + 2 k + 1)}

Step 2: Decompose the General Term into a Telescoping Form○ Expand

Notice that the numerator $4 k$ is the difference of the factors in the denominator: $(2 k^{2} + 2 k + 1) - (2 k^{2} - 2 k + 1) = 4 k$ . This allows us to express $T_{k}$ as a difference of two terms:

T_{k} = \frac{(2 k^{2} + 2 k + 1) - (2 k^{2} - 2 k + 1)}{(2 k^{2} - 2 k + 1) (2 k^{2} + 2 k + 1)} = \frac{1}{2 k^{2} - 2 k + 1} - \frac{1}{2 k^{2} + 2 k + 1}

Let $f (k) = \frac{1}{2 k^{2} - 2 k + 1}$ . Then $f (k + 1) = \frac{1}{2 (k + 1)^{2} - 2 (k + 1) + 1} = \frac{1}{2 (k^{2} + 2 k + 1) - 2 k - 2 + 1} = \frac{1}{2 k^{2} + 2 k + 1}$ . Thus, $T_{k} = f (k) - f (k + 1)$ , which is a telescoping form.

💡 Teacher's Secret Hint

Recognizing the difference in the numerator is key to forming the telescoping sum.

Step 3: Calculate the Sum of the First 10 Terms and Final Result○ Expand

The sum of the first 10 terms, $S_{10}$ , is given by:

S_{10} = \sum_{k = 1}^{10} T_{k} = \sum_{k = 1}^{10} (f (k) - f (k + 1))

This is a telescoping sum, so most terms cancel out, leaving:

S_{10} = f (1) - f (11)

Calculate $f (1)$ and $f (11)$ :

f (1) = \frac{1}{2 (1)^{2} - 2 (1) + 1} = \frac{1}{1} = 1

f (11) = \frac{1}{2 (11)^{2} - 2 (11) + 1} = \frac{1}{2 (121) - 22 + 1} = \frac{1}{242 - 22 + 1} = \frac{1}{221}

Substitute these values back into the sum:

S_{10} = 1 - \frac{1}{221} = \frac{221 - 1}{221} = \frac{220}{221}

We are given that the sum is $\frac{m}{n}$ where $gcd (m, n) = 1$ . Comparing with our result, $m = 220$ and $n = 221$ . We verify that $gcd (220, 221) = 1$ (since $221 = 220 + 1$ ). Finally, we need to find $m + n$ :

m + n = 220 + 221 = 441

💡 Teacher's Secret Hint

Ensure to correctly identify the first and last terms after the telescoping cancellation.

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Maths Question 22 – JEE-MAIN 2025

Clue 1 of 3: Series Term Structure

🧠 Full Solution Path