Multiple choice

The sum of the series $\dfrac {3}{1^{2}} + \dfrac {5}{1^{2} + 2^{2}} + \dfrac {7}{1^{2} +2^{2} + 3^{2}} + ....$ upto $n$ terms, is

  1. $\dfrac {n}{n + 1}$
  2. $\dfrac {n + 2}{n + 1}$
  3. $\dfrac {6n}{n + 1}$
  4. $\dfrac {6(n + 2)}{n + 1}$
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C Correct answer
Explanation

The n-th term is (2n+1) / (sum of first n squares). Sum of first n squares = n(n+1)(2n+1)/6. So n-th term = (2n+1) / [n(n+1)(2n+1)/6] = 6 / [n(n+1)] = 6 * (1/n - 1/(n+1)). The sum of this telescoping series is 6 * (1 - 1/(n+1)) = 6 * (n / (n+1)) = 6n / (n+1).

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