Question

The sum of the series \[ \frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \ldots \text{ up to 10 terms} \] is

• A. \( \frac{45}{109} \)
• B. \( -\frac{45}{109} \)
• C. \( \frac{55}{109} \)
• D. \( -\frac{55}{109} \)

Answer 1

The Correct Option is D

The task is to compute the sum of the following series, up to 10 terms:

\[ \frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \ldots \text{ for } n=1 \text{ to } 10 \]

The general term of the series is given by \( \frac{n}{1 - 3n^2 + n^4} \).

We first evaluate the denominator \( 1 - 3n^2 + n^4 \) for \( n \) from 1 to 10:

• For \(n = 1\): \(1 - 3(1)^2 + (1)^4 = 1 - 3 + 1 = -1\)
• For \(n = 2\): \(1 - 3(2)^2 + (2)^4 = 1 - 12 + 16 = 5\)
• For \(n = 3\): \(1 - 3(3)^2 + (3)^4 = 1 - 27 + 81 = 55\)
• For \(n = 4\): \(1 - 3(4)^2 + (4)^4 = 1 - 48 + 256 = 209\)
• For \(n = 5\): \(1 - 3(5)^2 + (5)^4 = 1 - 75 + 625 = 551\)
• For \(n = 6\): \(1 - 3(6)^2 + (6)^4 = 1 - 108 + 1296 = 1189\)
• For \(n = 7\): \(1 - 3(7)^2 + (7)^4 = 1 - 147 + 2401 = 2255\)
• For \(n = 8\): \(1 - 3(8)^2 + (8)^4 = 1 - 192 + 4096 = 3905\)
• For \(n = 9\): \(1 - 3(9)^2 + (9)^4 = 1 - 243 + 6561 = 6319\)
• For \(n = 10\): \(1 - 3(10)^2 + (10)^4 = 1 - 300 + 10000 = 9701\)

The series then becomes:

\[ \frac{1}{-1} + \frac{2}{5} + \frac{3}{55} + \frac{4}{209} + \frac{5}{551} + \frac{6}{1189} + \frac{7}{2255} + \frac{8}{3905} + \frac{9}{6319} + \frac{10}{9701} \]

The sum of these terms is represented as:

\[ S = -1 + \frac{2}{5} + \frac{3}{55} + \frac{4}{209} + \frac{5}{551} + \frac{6}{1189} + \frac{7}{2255} + \frac{8}{3905} + \frac{9}{6319} + \frac{10}{9701} \]

Upon accurate calculation of this sum, the result is approximately \( -\frac{55}{109} \).

The final answer is: \(-\frac{55}{109}\)