Question

The sum of the series
\(∑^∞_{ n=1} \frac{1}{(4n-3)(4n+1) }\) 
is equal to_____.(Round off to two decimal places)

Answer 1

Correct Answer: 0.24 - 0.26

To find the sum of the series \(∑^∞_{n=1} \frac{1}{(4n-3)(4n+1)}\), we start by simplifying the general term using partial fraction decomposition. Expressing \(\frac{1}{(4n-3)(4n+1)}\) as a sum of fractions:

\(\frac{1}{(4n-3)(4n+1)} = \frac{A}{4n-3} + \frac{B}{4n+1}\).
Multiplying through by the denominator \((4n-3)(4n+1)\), we obtain:

\[1 = A(4n+1) + B(4n-3)\]

Equating coefficients gives a system of equations:

• \(4A + 4B = 0\)
• \(A - 3B = 1\)

Solving these, we find:

\(A = \frac{1}{4}\), \(B = -\frac{1}{4}\).

Thus, \(\frac{1}{(4n-3)(4n+1)} = \frac{1}{4}\left(\frac{1}{4n-3} - \frac{1}{4n+1}\right)\).

The series becomes:

\(\frac{1}{4}\left(\sum_{n=1}^{\infty} \left(\frac{1}{4n-3} - \frac{1}{4n+1}\right)\right)\)

This is a telescoping series, where intermediate terms cancel out, leaving:

\(\frac{1}{4}\left(1 - \lim_{n \to \infty} \frac{1}{4n+1}\right) = \frac{1}{4}(1 - 0) = \frac{1}{4}\)

Converting \(\frac{1}{4}\) to decimal form gives \(0.25\).

Thus, the sum of the series is 0.25, rounded to two decimal places.