Multiple choice

The sum to $n$ terms of the series: $\dfrac { 1 }{ 1+{ 1 }^{ 2 }+{ 1 }^{ 4 } } +\dfrac { 2 }{ 1+{ 2 }^{ 2 }+{ 2 }^{ 4 } } +\dfrac { 3 }{ 1+{ 3 }^{ 2 }+{ 3 }^{ 4 } } +\cdots $ is

  1. $\dfrac { { n }^{ 2 }+1 }{ 2\left( { n }^{ 2 }+n+1 \right) } $
  2. $\dfrac { { n }^{ 2 }+n }{ \left( { n }^{ 2 }+n+1 \right) } $
  3. $\dfrac { { n }^{ 2 }+n }{ 2\left( { n }^{ 2 }+n+1 \right) } $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The general term is n / (1 + n^2 + n^4). Note that 1 + n^2 + n^4 = (n^2 + n + 1)(n^2 - n + 1). Using partial fractions, the sum telescopes to (n^2 + n) / (2(n^2 + n + 1)).

AI explanation

Factor the denominator of the nth term as (1 + t^2 + t^4) = (t^2 + t + 1)(t^2 - t + 1) by treating it as a quadratic in t^2. This allows you to rewrite the term as 1/2 multiplied by the difference [(t^2 + t) - (t^2 - t)] divided by the product of those factors. Using partial fractions, the nth term simplifies to 1/2[(1/(t^2 - t + 1)) - (1/(t^2 + t + 1))], which shifts to 1/2[(1/(t^2 - t + 1)) - (1/((t+1)^2 - (t+1) + 1))]. When summed to n terms, the series telescopes to 1/2[1 - 1/(n^2 + n + 1)], leaving the final result as (n^2 + n) / (2(n^2 + n + 1)).

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