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 problem requires us to find the sum of a series given by:

\[ \frac{1}{1 - 3 \cdot 1^2 + 1^4} + \frac{2}{1 - 3 \cdot 2^2 + 2^4} + \frac{3}{1 - 3 \cdot 3^2 + 3^4} + \ldots \text{ up to 10 terms} \]

Let's start by simplifying each term in the series: 

1. General term of the series can be expressed as \( \frac{n}{1 - 3n^2 + n^4} \).
2. We evaluate the denominator \( 1 - 3n^2 + n^4 \) for each term individually.

Substituting \(n = 1\), \(n = 2\), ..., up to \(n = 10\), we get:

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

Now, the series becomes:

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

To find the sum of these terms, add them accordingly:

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

Calculation leads to an approximate result for the sum:

• Adding these fractions accurately shows that the sum is almost \( -\frac{55}{109} \).

Therefore, the correct answer is: \(-\frac{55}{109}\)

Answer 2

General term of the sequence,

$T_r = \frac{r}{1 - 3r^2 + r^4}$

$T_r = \frac{r}{r^4 - 3r^2 + 1 - r^2}$

$T_r = \frac{r}{(r^2 - 1)^2 - r^2}$

$T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}$

$T_r = \frac{1}{2} \left[ \frac{1}{(r^2 - r - 1)} - \frac{1}{(r^2 + r - 1)} \right]$

Sum of 10 terms,

$\sum_{r=1}^{10} T_r = \frac{1}{2} \left[ \frac{1}{-1} - \frac{1}{109} \right] = \frac{55}{109}$