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

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q5 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #397

If the sum of the first 10 terms of the series

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

\frac{m}{n}

gcd (m, n) = 1

, then

m + n

is equal to :

Clue 1 of 3: Series Recognition

Identify the general term of the series and look for a pattern that allows for a telescoping sum.

🧠 Full Solution Path

Step 1: Identify the general term and factorize the denominator✦ Active

The general term of the series is $T_{k} = \frac{k}{1 + 4 k^{4}}$ . The denominator $1 + 4 k^{4}$ can be factored using the identity $a^{4} + 4 b^{4} = (a^{2} + 2 b^{2} + 2 a b) (a^{2} + 2 b^{2} - 2 a b)$ or by completing the square:

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

This is a difference of squares:

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

So, the general term is:

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

Step 2: Express the general term as a difference of two terms○ Expand

Let $A = 2 k^{2} - 2 k + 1$ and $B = 2 k^{2} + 2 k + 1$ . Then $B - A = (2 k^{2} + 2 k + 1) - (2 k^{2} - 2 k + 1) = 4 k$ . So, $k = \frac{1}{4} (B - A)$ . Substituting this into $T_{k}$ :

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

Let $g (k) = \frac{1}{2 k^{2} - 2 k + 1}$ . Then $g (k + 1)$ is:

g (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, the general term can be written in a telescoping form:

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

💡 Teacher's Secret Hint

This transformation is key for telescoping sums. Ensure the difference in the numerator matches the difference of the denominators' factors.

Step 3: Calculate the sum of the first 10 terms and find

m + n

○ Expand

The sum of the first 10 terms is $S_{10} = \sum_{k = 1}^{10} T_{k} = \frac{1}{4} \sum_{k = 1}^{10} (g (k) - g (k + 1))$ . This is a telescoping sum:

S_{10} = \frac{1}{4} [(g (1) - g (2)) + (g (2) - g (3)) + \dots + (g (10) - g (11))] = \frac{1}{4} (g (1) - g (11))

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

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

g (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 into $S_{10}$ :

S_{10} = \frac{1}{4} (1 - \frac{1}{221}) = \frac{1}{4} (\frac{221 - 1}{221}) = \frac{1}{4} (\frac{220}{221}) = \frac{55}{221}

The sum is $\frac{m}{n} = \frac{55}{221}$ . We check $gcd (55, 221)$ : $55 = 5 \times 11$ and $221 = 13 \times 17$ . They are coprime, so $gcd (55, 221) = 1$ . Therefore, $m = 55$ and $n = 221$ . The required value is $m + n = 55 + 221 = 276$ .

💡 Teacher's Secret Hint

Always simplify the fraction to its lowest terms to ensure $gcd (m, n) = 1$ before calculating $m + n$ .

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Series Recognition

🧠 Full Solution Path