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 determine the sum to \( n \) terms of the series \(\frac{3}{1^2} + \frac{5}{1^2+2^2} + \frac{7}{1^2+2^2+3^2} + \ldots\), let's analyze the pattern and formulae behind the series.

The general term of the series is given by:

\(T_k = \frac{2k + 1}{1^2 + 2^2 + \ldots + k^2}\)

The denominator is the sum of squares of the first \( k \) natural numbers. This is given by the formula:

\(1^2 + 2^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6}\)

Thus, the term \(T_k\) becomes:

\(T_k = \frac{2k + 1}{\frac{k(k+1)(2k+1)}{6}} = \frac{6(2k + 1)}{k(k+1)(2k+1)}\)

Simplifying further by canceling \((2k + 1)\) from the numerator and denominator, we get:

\(T_k = \frac{6}{k(k+1)}\)

The sum of the first \( n \) terms of the series \( S_n \) is:

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

Using the property of partial fractions, we can write:

\(\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\)

Therefore, the sum becomes:

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

This simplifies to:

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

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

\(S_n = \frac{6n}{n+1}\)

Conclusion: The correct answer is \(\frac{6n}{n+1}\).