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JEE Mathematics 2025 Question with Solution

If the sum of the first 1010 terms of the series 4⋅11+4⋅14+4⋅21+4⋅24+4⋅31+4⋅34+⋯\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} + \cdots is mn\frac{m}{n}, where gcd⁡(m,n)=1\gcd(m, n) = 1, then m+nm + n is equal to _____

  • A

    1515

  • B

    2424

  • C

    4141

  • D

    7676

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: We need the sum of the first 1010 terms of the series

S10=∑n=1104n1+4n4.S_{10}=\sum_{n=1}^{10}\frac{4n}{1+4n^{4}}.

Find: If S10=mnS_{10}=\frac{m}{n} in lowest terms, find m+nm+n.

Factor the denominator:

1+4n4=(2n2−2n+1)(2n2+2n+1).1+4n^{4}=(2n^{2}-2n+1)(2n^{2}+2n+1).

So,

4n1+4n4=4n(2n2−2n+1)(2n2+2n+1)=12n2−2n+1−12n2+2n+1.\frac{4n}{1+4n^{4}}=\frac{4n}{(2n^{2}-2n+1)(2n^{2}+2n+1)}=\frac{1}{2n^{2}-2n+1}-\frac{1}{2n^{2}+2n+1}.

Therefore,

S10=∑n=110(12n2−2n+1−12n2+2n+1).S_{10}=\sum_{n=1}^{10}\left(\frac{1}{2n^{2}-2n+1}-\frac{1}{2n^{2}+2n+1}\right).

This is a telescoping series because

2(n+1)2−2(n+1)+1=2n2+2n+1.2(n+1)^{2}-2(n+1)+1=2n^{2}+2n+1.

After cancellation, only the first positive term and the last negative term remain:

S10=12⋅12−2⋅1+1−12⋅102+2⋅10+1=1−1221=220221.S_{10}=\frac{1}{2\cdot 1^{2}-2\cdot 1+1}-\frac{1}{2\cdot 10^{2}+2\cdot 10+1}=1-\frac{1}{221}=\frac{220}{221}.

Thus m=220m=220 and n=221n=221, so

m+n=220+221=441.m+n=220+221=441.

Therefore, the working in the solution gives 441441, which does not match any listed option. The source options show D = 7676, but the extracted solution concludes 441441.

Hence, based on the provided options, the marked answer is D, while the solution concludes 441441.

Telescoping Expansion

Given:

Tn=4n1+4n4.T_n=\frac{4n}{1+4n^4}.

Find: Sum of the first 1010 terms and then compute m+nm+n.

Using

4n1+4n4=12n2−2n+1−12n2+2n+1,\frac{4n}{1+4n^{4}}=\frac{1}{2n^{2}-2n+1}-\frac{1}{2n^{2}+2n+1},

we get

S10=(11−15)+(15−113)+(113−125)+⋯+(1181−1221).\begin{aligned} S_{10} &= \left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{25}\right)+\cdots \\ &\quad +\left(\frac{1}{181}-\frac{1}{221}\right). \end{aligned}

All intermediate terms cancel, so

S10=1−1221=220221.S_{10}=1-\frac{1}{221}=\frac{220}{221}.

Hence,

m+n=220+221=441.m+n=220+221=441.

Therefore, the numerical result from the solution is 441441.

Common mistakes

  • A common mistake is to treat the denominator 1+4n41+4n^4 as if it does not factor. That prevents spotting the telescoping pattern. Instead, factor it as (2n2−2n+1)(2n2+2n+1)(2n^2-2n+1)(2n^2+2n+1) first.

  • Another mistake is to miss the index shift in telescoping. The term 2n2+2n+12n^2+2n+1 matches 2(n+1)2−2(n+1)+12(n+1)^2-2(n+1)+1, so cancellation happens between consecutive terms, not within the same term.

  • Students may compute S10S_{10} correctly as 220221\frac{220}{221} but then forget that the question asks for m+nm+n, not the sum itself. After identifying m=220m=220 and n=221n=221, add them.

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