Multiple choice

The sum of series $\displaystyle \frac {1^2}{1}+\frac {1^2+2^2}{1+2}+\frac {1^2+2^2+3^2}{1+2+3}+....$ upto n terms is

  1. $\displaystyle\frac {1}{3}(2n+1)$
  2. $\displaystyle\frac {1}{3}n^2$
  3. $\displaystyle\frac {1}{3}(n+2)$
  4. $\displaystyle\frac {1}{3}n(n+2)$
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D Correct answer
Explanation

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

AI explanation

Using the formulas for the sum of squares and the sum of the first n natural numbers, the numerator is n(n+1)(2n+1)/6 and the denominator is n(n+1)/2. The kth term of the series simplifies to (2k+1)/3. To find the total sum to n terms, we evaluate the series (3 + 5 + 7 + ... up to n terms). The sum of this arithmetic progression is n/2 multiplied by 2 times 3 plus (n-1) times 2, which equals n(n+2). Dividing by 3 gives the result of n(n+2)/3.

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