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MCQMediumJEE 2025Sum of Series

JEE Mathematics 2025 Question with Solution

If the sum of the first 2020 terms of the series 4⋅14+3⋅12+14+4⋅24+3⋅22+24+4⋅34+3⋅32+34+4⋅44+3⋅42+44+…\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 is mn\frac{m}{n}, where mm and nn are coprime, then m+nm + n is equal to:

  • A

    423423

  • B

    420420

  • C

    421421

  • D

    422422

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

S20=∑r=1204r4+3r2+r4S_{20} = \sum_{r=1}^{20} \frac{4r}{4 + 3r^2 + r^4}

Find: The value of m+nm+n when S20=mnS_{20} = \frac{m}{n} in lowest terms.

Factor the denominator:

r4+3r2+4=(r2−r+2)(r2+r+2)r^4 + 3r^2 + 4 = (r^2-r+2)(r^2+r+2)

So,

4r4+3r2+r4=4r(r2−r+2)(r2+r+2)\frac{4r}{4 + 3r^2 + r^4} = \frac{4r}{(r^2-r+2)(r^2+r+2)}

Now write it in telescoping form:

4r(r2−r+2)(r2+r+2)=2(1r2−r+2−1r2+r+2)\frac{4r}{(r^2-r+2)(r^2+r+2)} = 2\left(\frac{1}{r^2-r+2} - \frac{1}{r^2+r+2}\right)

Since

r2+r+2=(r+1)2−(r+1)+2r^2+r+2 = (r+1)^2-(r+1)+2

let

ar=1r2−r+2a_r = \frac{1}{r^2-r+2}

Then the series becomes

S20=2∑r=120(ar−ar+1)S_{20} = 2\sum_{r=1}^{20}(a_r-a_{r+1})

Hence the sum telescopes:

S20=2(a1−a21)=2(112−1+2−1212−21+2)S_{20} = 2(a_1-a_{21}) = 2\left(\frac{1}{1^2-1+2} - \frac{1}{21^2-21+2}\right) =2(12−1422)= 2\left(\frac{1}{2} - \frac{1}{422}\right) =1−2422=1−1211=210211= 1 - \frac{2}{422} = 1 - \frac{1}{211} = \frac{210}{211}

Therefore, m=210m=210 and n=211n=211, so

m+n=210+211=421m+n = 210+211 = 421

The correct option is C.

Using the general partial sum

From the telescoping form,

Sn=2∑r=1n(1r2−r+2−1r2+r+2)S_n = 2\sum_{r=1}^{n}\left(\frac{1}{r^2-r+2} - \frac{1}{r^2+r+2}\right)

This gives

Sn=2(12−1n2+n+2)S_n = 2\left(\frac{1}{2} - \frac{1}{n^2+n+2}\right)

because r2+r+2r^2+r+2 in one term matches the next denominator (r+1)2−(r+1)+2(r+1)^2-(r+1)+2.

So,

Sn=1−2n2+n+2=n2+nn2+n+2S_n = 1 - \frac{2}{n^2+n+2} = \frac{n^2+n}{n^2+n+2}

For n=20n=20,

S20=202+20202+20+2=420422=210211S_{20} = \frac{20^2+20}{20^2+20+2} = \frac{420}{422} = \frac{210}{211}

Thus, m=210m=210 and n=211n=211.

Therefore,

m+n=421m+n = 421

So the correct option is C.

Common mistakes

  • Factoring r4+3r2+4r^4+3r^2+4 incorrectly. This breaks the telescoping structure. Use r4+3r2+4=(r2−r+2)(r2+r+2)r^4+3r^2+4=(r^2-r+2)(r^2+r+2) before splitting the term.

  • Stopping at 420422\frac{420}{422} and taking m=420,n=422m=420, n=422 directly. The fraction must be reduced to coprime form first, so 420422=210211\frac{420}{422}=\frac{210}{211}.

  • Using the misleading solution statement S=4211S=\frac{421}{1}. The series sum is not 421421; rather, m+n=421m+n=421 after writing the sum as 210211\frac{210}{211}.

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