Question

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

• A. \(\frac{6n}{n+1}\)
• B. \(\frac{9n}{n+1}\)
• C. \(\frac{12n}{n+1}\)
• D. \(\frac{3n}{n+1}\)

Hint:

Telescoping series: \(\sum \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1}\).

Answer 1

The Correct Option is A

To solve the problem, we need to determine the sum to \( n \) terms of the given series: 

\[ \frac{3}{1^2} + \frac{5}{1^2+2^2} + \frac{7}{1^2+2^2+3^2} + \cdots \]

This series can be generalized for the \( n \)-th term as follows:

\[ a_n = \frac{2n+1}{1^2 + 2^2 + 3^2 + \cdots + n^2} \]

The denominator, \( 1^2 + 2^2 + 3^2 + \cdots + n^2 \), is the sum of squares of the first \( n \) natural numbers, which is given by the formula:

\[ \text{Sum} = \frac{n(n+1)(2n+1)}{6} \]

Thus, the \( n \)-th term can be rewritten as:

\[ a_n = \frac{2n+1}{\frac{n(n+1)(2n+1)}{6}} = \frac{6(2n+1)}{n(n+1)(2n+1)} \]

Cancel the common term \( 2n+1 \) from numerator and denominator:

\[ a_n = \frac{6}{n(n+1)} \]

Now, summing these terms up to \( n \) terms gives:

\[ S_n = \sum_{k=1}^{n} \frac{6}{k(k+1)} \]

Using partial fraction decomposition:

\[ \frac{6}{k(k+1)} = 6 \left( \frac{1}{k} - \frac{1}{k+1} \right) \]

The series becomes a telescoping series:

\[ S_n = 6 \left( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \right) \]

All terms cancel out except the first and the last one:

\[ S_n = 6 \left( 1 - \frac{1}{n+1} \right) = 6 \left( \frac{n+1-1}{n+1} \right) = 6 \left( \frac{n}{n+1} \right) \]

Thus, the sum to \( n \) terms of the series is:

\[ S_n = \frac{6n}{n+1} \]

This confirms that the correct answer is \(\frac{6n}{n+1}\).