Multiple choice

The sum of the first n terms of the series ${ 1 }^{ 2 }+{ 2.2 }^{ 2 }+{ 3 }^{ 2 }+{ 2.4 }^{ 2 }+{ 5 }^{ 2 }+{ 2.6 }^{ 2 }+.....$ is $\cfrac { n{ \left( n+1 \right) }^{ 2 } }{ 2 } $, when n is even. When n is odd, the sum is-

  1. $\cfrac { { n }^{ 2 }(n+1) }{ 2 } $
  2. $\cfrac { n(n+1)(2n+1) }{ 6 } $
  3. $\cfrac { n{ (n+1) }^{ 2 } }{ 2 } $
  4. $\cfrac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 2 } $
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A Correct answer
Explanation

The series is 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2... For odd n, the sum follows the pattern n^2(n+1)/2.

AI explanation

For an odd number of terms n, the sum equals the sum of the first (n-1) terms plus the nth term. Given the sum of (n-1) even terms is (n-1)(n)^2 / 2 and the nth term of the series is n^2, add these two expressions. This results in (n^3 - n^2)/2 + n^2, which simplifies to (n^3 - n^2 + 2n^2)/2, yielding the final odd sum formula n^2(n+1)/2.

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