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Sequences and Series question

2025 · 4 Apr · Shift 2 · Q33
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  5. /2025 · 4 Apr · Shift 2 · Q33

Sequences and Series question

2025 · 4 Apr · Shift 2 · Q33

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If the sum of the first 20 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}+\ldots \cdot4+3⋅12+144⋅1​+4+3⋅22+244⋅2​+4+3⋅32+344⋅3​+4+3⋅42+444⋅4​+…⋅ is mn\frac{\mathrm{m}}{\mathrm{n}}nm​, where m and n are coprime, then m+n\mathrm{m}+\mathrm{n}m+n is equal to :
  1. A
    423
  2. B
    421
  3. C
    422
  4. D
    420
View written solutionFree

Correct answer: B

  1. Write the general term

The given series has general term

Tr=4r4+3r2+r4.T_r=\frac{4r}{4+3r^2+r^4}.Tr​=4+3r2+r44r​.

So we need

S20=∑r=1204rr4+3r2+4.S_{20}=\sum_{r=1}^{20} \frac{4r}{r^4+3r^2+4}.S20​=∑r=120​r4+3r2+44r​.

  1. Factor the denominator

Observe that

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

Indeed,

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

Hence

Tr=4r(r2−r+2)(r2+r+2).T_r=\frac{4r}{(r^2-r+2)(r^2+r+2)}.Tr​=(r2−r+2)(r2+r+2)4r​.

  1. Convert into telescoping form

Notice that

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

Therefore,

1r2−r+2−1r2+r+2=(r2+r+2)−(r2−r+2)(r2−r+2)(r2+r+2)=2r(r2−r+2)(r2+r+2).\frac{1}{r^2-r+2}-\frac{1}{r^2+r+2} =\frac{(r^2+r+2)-(r^2-r+2)}{(r^2-r+2)(r^2+r+2)} =\frac{2r}{(r^2-r+2)(r^2+r+2)}.r2−r+21​−r2+r+21​=(r2−r+2)(r2+r+2)(r2+r+2)−(r2−r+2)​=(r2−r+2)(r2+r+2)2r​.

Multiplying by 222,

Tr=2(1r2−r+2−1r2+r+2).T_r=2\left(\frac{1}{r^2-r+2}-\frac{1}{r^2+r+2}\right).Tr​=2(r2−r+21​−r2+r+21​).

Now simplify the quadratic expressions:

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

Also notice

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

so if we define

ar=r2−r+2,a_r=r^2-r+2,ar​=r2−r+2, then

ar+1=r2+r+2.a_{r+1}=r^2+r+2.ar+1​=r2+r+2.

Thus

Tr=2(1ar−1ar+1),T_r=2\left(\frac{1}{a_r}-\frac{1}{a_{r+1}}\right),Tr​=2(ar​1​−ar+1​1​), which is telescoping.

  1. Sum the first 20 terms

So

S20=2∑r=120(1ar−1ar+1)=2(1a1−1a21).S_{20}=2\sum_{r=1}^{20}\left(\frac{1}{a_r}-\frac{1}{a_{r+1}}\right) =2\left(\frac{1}{a_1}-\frac{1}{a_{21}}\right).S20​=2r=1∑20​(ar​1​−ar+1​1​)=2(a1​1​−a21​1​).

Now,

a1=12−1+2=2,a_1=1^2-1+2=2,a1​=12−1+2=2,

and

a21=212−21+2=441−21+2=422.a_{21}=21^2-21+2=441-21+2=422.a21​=212−21+2=441−21+2=422.

Therefore,

S20=2(12−1422)=1−2422=1−1211=210211.S_{20}=2\left(\frac{1}{2}-\frac{1}{422}\right) =1-\frac{2}{422} =1-\frac{1}{211} =\frac{210}{211}.S20​=2(21​−4221​)=1−4222​=1−2111​=211210​.

Thus,

m=210,n=211.m=210,\quad n=211.m=210,n=211.

So,

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

  1. Check options
  • A: 423423423
  • B: 421421421 ✅
  • C: 422422422
  • D: 420420420

Hence the correct option is B.

  1. Comparison with stored answer

Stored correct answer = B.

Our derived answer is also B.

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