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Sequences and Series question

2025 · 2 Apr · Shift 2 · Q48
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Sequences and Series question

2025 · 2 Apr · Shift 2 · Q48

JEE MainMathematicsSequences and SeriesNumerical+4 / −1
If the sum of the first 10 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}+\ldots .1+4⋅144⋅1​+1+4⋅244⋅2​+1+4⋅344⋅3​+….. is mn\frac{\mathrm{m}}{\mathrm{n}}nm​, where gcd⁡(m,n)=1\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1gcd(m,n)=1, then m+n\mathrm{m}+\mathrm{n}m+n is equal to ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 441

  1. The given series has general term Tr=4r1+4r4T_r=\frac{4r}{1+4r^4}Tr​=1+4r44r​ and we need S10=∑r=1104r1+4r4.S_{10}=\sum_{r=1}^{10}\frac{4r}{1+4r^4}. S10​=∑r=110​1+4r44r​.

  2. Factor the denominator: 1+4r4=(2r2−2r+1)(2r2+2r+1).1+4r^4=(2r^2-2r+1)(2r^2+2r+1).1+4r4=(2r2−2r+1)(2r2+2r+1).

    This is because (2r2−2r+1)(2r2+2r+1)=((2r2+1)−2r)((2r2+1)+2r)=(2r2+1)2−(2r)2=4r4+1.(2r^2-2r+1)(2r^2+2r+1)=((2r^2+1)-2r)((2r^2+1)+2r)=(2r^2+1)^2-(2r)^2=4r^4+1.(2r2−2r+1)(2r2+2r+1)=((2r2+1)−2r)((2r2+1)+2r)=(2r2+1)2−(2r)2=4r4+1.

  3. Now observe:

    =\frac{(2r^2+2r+1)-(2r^2-2r+1)}{(2r^2-2r+1)(2r^2+2r+1)}$$ $$=\frac{4r}{1+4r^4}. $$ Hence, $$T_r=\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}. $$

  4. Rewrite each denominator: 2r2−2r+1=2(r−1/2)2+1/2,2r^2-2r+1=2(r-1/2)^2+1/2,2r2−2r+1=2(r−1/2)2+1/2, but more importantly, 2r2−2r+1=2(r−1)2+2(r−1)+1.2r^2-2r+1=2(r-1)^2+2(r-1)+1.2r2−2r+1=2(r−1)2+2(r−1)+1.

    Notice that 2r2+2r+1=2(r+1)2−2(r+1)+1.2r^2+2r+1=2(r+1)^2-2(r+1)+1.2r2+2r+1=2(r+1)2−2(r+1)+1.

    So if we define Ar=12r2+2r+1,A_r=\frac{1}{2r^2+2r+1},Ar​=2r2+2r+11​, then 12r2−2r+1=Ar−1.\frac{1}{2r^2-2r+1}=A_{r-1}. 2r2−2r+11​=Ar−1​.

    Therefore, Tr=Ar−1−Ar.T_r=A_{r-1}-A_r. Tr​=Ar−1​−Ar​.

  5. Hence the sum telescopes:

    =A_0-A_{10}. $$ Now, $$A_0=\frac{1}{1}=1,$$ and $$A_{10}=\frac{1}{2(10)^2+2(10)+1}=\frac{1}{200+20+1}=\frac{1}{221}. $$ Therefore, $$S_{10}=1-\frac{1}{221}=\frac{220}{221}. $$

  6. Thus, m=220,n=221,m=220,\quad n=221,m=220,n=221, with gcd⁡(220,221)=1.\gcd(220,221)=1.gcd(220,221)=1.

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

  7. Comparison with stored answer: The stored correct answer is 441441441, which matches our result.

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