Multiple choice

The sum to $n$ terms of the series $\displaystyle \frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...$ is

  1. $\displaystyle \frac{6n}{n+1}$
  2. $\displaystyle \frac{9n}{n+1}$
  3. $\displaystyle \frac{12n}{n+1}$
  4. $\displaystyle \frac{3n}{n+1}$
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A Correct answer
Explanation

The nth term is (2n+1) / (sum of first n squares). The sum of first n squares is n(n+1)(2n+1)/6. Thus, the nth term is (2n+1) / (n(n+1)(2n+1)/6) = 6 / (n(n+1)) = 6(1/n - 1/(n+1)). Summing this telescoping series gives 6(1 - 1/(n+1)) = 6n/(n+1).

AI explanation

The denominator of the nth term represents the sum of the first n squares, calculated using the formula n(n+1)(2n+1)/6. The numerator of the nth term forms an arithmetic progression given by 2n + 1, so the nth term simplifies to (2n+1) divided by n(n+1)(2n+1)/6, which reduces to 6/(n(n+1)). By the method of differences, this term can be split into 6/n - 6/(n+1), making the sum of n terms telescope to 6 - 6/(n+1), which equals 6n/(n+1).

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