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JEE Main 2020 · Mathematics · Sequences and Series · hard

The value of Σk=1^10(k)/(k^4+k^2+1) is

Single correct option

The value of ∑k=110kk4+k2+1\displaystyle\sum_{k=1}^{10}\dfrac{k}{k^4+k^2+1} is
  1. A. 1021\dfrac{10}{21}
  2. B. 110111\dfrac{110}{111}
  3. C. 55111\dfrac{55}{111}
  4. D. 511\dfrac{5}{11}
Show answer and solution

Answer: (C) 55111\dfrac{55}{111}

Factor the denominator: k4+k2+1=(k2+k+1)(k2−k+1)k^4+k^2+1=(k^2+k+1)(k^2-k+1). Since (k2+k+1)−(k2−k+1)=2k(k^2+k+1)-(k^2-k+1)=2k, we get kk4+k2+1=12[1k2−k+1−1k2+k+1].\frac{k}{k^4+k^2+1}=\frac12\left[\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right]. Because (k+1)2−(k+1)+1=k2+k+1(k+1)^2-(k+1)+1=k^2+k+1, the sum telescopes: ∑k=1n=12[1−1n2+n+1]=n(n+1)2(n2+n+1).\sum_{k=1}^{n}=\frac12\left[1-\frac{1}{n^2+n+1}\right]=\frac{n(n+1)}{2(n^2+n+1)}. For n=10n=10 this is 110222=55111\dfrac{110}{222}=\dfrac{55}{111}. A numerical check of the first few terms (13,221,…\tfrac13,\tfrac{2}{21},\ldots) agrees. Check with n=1n=1: the formula gives 26=13\dfrac{2}{6}=\dfrac13, which equals the first term 11+1+1\dfrac{1}{1+1+1}.
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