If the sum of the first 20 terms of the series 4*1/(4+3*1^2+1^4) + 4*2/(4+3*2^2+2^4) + 4*3/(4+3*3^2+3^4) + 4*4/(4+3*4^2+4^4) + .... is m/n, where m and n are coprime, then m+n is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q5 – JEE-MAIN 2025

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Home Maths Sequences and SeriesQuestion #410

If the sum of the first 20 terms of the series

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

\frac{m}{n}

, where m and n are coprime, then m+n is equal to :

Clue 1 of 3: Series Analysis

Examine the general term of the series to identify its structure and potential for simplification.

🧠 Full Solution Path

Step 1: Simplify the general term✦ Active

The general term of the series is $T_{k} = \frac{4 k}{4 + 3 k^{2} + k^{4}}$ . The denominator can be factored by completing the square:

k^{4} + 3 k^{2} + 4 = (k^{4} + 4 k^{2} + 4) - k^{2} = (k^{2} + 2)^{2} - k^{2} = (k^{2} - k + 2) (k^{2} + k + 2)

So, the general term becomes:

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

Step 2: Express

T_{k}

as a difference of two terms (telescopic form)○ Expand

Let $A = k^{2} - k + 2$ and $B = k^{2} + k + 2$ . Notice that $B - A = (k^{2} + k + 2) - (k^{2} - k + 2) = 2 k$ . The numerator is $4 k = 2 (2 k) = 2 (B - A)$ . Thus, $T_{k}$ can be written as:

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

Let $f (k) = \frac{1}{k^{2} - k + 2}$ . Then $f (k + 1) = \frac{1}{(k + 1)^{2} - (k + 1) + 2} = \frac{1}{k^{2} + 2 k + 1 - k - 1 + 2} = \frac{1}{k^{2} + k + 2}$ . Therefore, $T_{k} = 2 (f (k) - f (k + 1))$ . This is a telescopic form.

💡 Teacher's Secret Hint

Recognizing the difference of terms in the numerator is key to forming a telescopic series.

Step 3: Calculate the sum and find

m + n

○ Expand

The sum of the first 20 terms is $S_{20} = \sum_{k = 1}^{20} T_{k} = \sum_{k = 1}^{20} 2 (f (k) - f (k + 1))$ . This is a telescopic sum, so:

S_{20} = 2 (f (1) - f (21))

Calculate $f (1)$ and $f (21)$ :

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

f (21) = \frac{1}{21^{2} - 21 + 2} = \frac{1}{441 - 21 + 2} = \frac{1}{422}

Substitute these values into the sum:

S_{20} = 2 (\frac{1}{2} - \frac{1}{422}) = 2 (\frac{211 - 1}{422}) = 2 (\frac{210}{422}) = \frac{420}{422} = \frac{210}{211}

The sum is $\frac{m}{n} = \frac{210}{211}$ . Since 211 is a prime number, $m = 210$ and $n = 211$ are coprime. We need to find $m + n$ :

m + n = 210 + 211 = 421

💡 Teacher's Secret Hint

Ensure the final fraction is in its simplest form to correctly identify coprime $m$ and $n$ .

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Maths Question 5 – JEE-MAIN 2025

Clue 1 of 3: Series Analysis

🧠 Full Solution Path