Multiple choice

The sum of $1$ st $n$ terms of the series $\cfrac { { 1 }^{ 2 } }{ 1 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +........$

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

The n-th term is (Sum of first n squares) / (Sum of first n integers) = [n(n+1)(2n+1)/6] / [n(n+1)/2] = (2n+1)/3. Summing this from 1 to n: (1/3) * [2 * n(n+1)/2 + n] = (1/3) * [n^2 + n + n] = n(n+2)/3.

AI explanation

We use the sum of squares formula and the sum of first n natural numbers formula to find the nth term of the given series. The nth term is (1^2 + 2^2 + ... + n^2) / (1 + 2 + ... + n), which simplifies to [n(n+1)(2n+1)/6] / [n(n+1)/2] and further reduces to (2n+1)/3. The sum of the first n terms of this sequence is the sum of (2k+1)/3 for k from 1 to n, which equals (1/3) * [2 * n(n+1)/2 + n] and simplifies to n(n+2)/3. The result is n(n+2)/3.

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