Sum of the n Terms of the Sequence

We are given the sequence:

\[ \sum_{k=1}^{n} \frac{k}{1 + k^2 + k^4} \]

Step-by-Step Solution:

We start by simplifying the denominator:

\[ 1 + k^2 + k^4 = k^4 + k^2 + 1 = (k^2 + k + 1)(k^2 - k + 1) \] So the term becomes: \[ \frac{k}{(k^2 + k + 1)(k^2 - k + 1)} \]

We now use partial fraction or telescoping logic:

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

Hence, the full sum becomes:

\[ \sum_{k=1}^{n} \frac{k}{1 + k^2 + k^4} = \frac{1}{2} \sum_{k=1}^{n} \left( \frac{1}{k^2 - k + 1} - \frac{1}{k^2 + k + 1} \right) \]

This is a telescoping series. Writing the first few terms:

\[ = \frac{1}{2} \left[ \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{13} \right) + \cdots \right] \]

All intermediate terms cancel out. We're left with:

\[ \boxed{ \sum_{k=1}^{n} \frac{k}{1 + k^2 + k^4} = \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right) } \]

Practice Questions

Q1. Find the value of \[ \sum_{k=1}^{4} \frac{k}{1 + k^2 + k^4} \]

Click to view solution

Using the formula:

\[ \sum_{k=1}^{n} \frac{k}{1 + k^2 + k^4} = \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right) \]

For \( n = 4 \):

\[ = \frac{1}{2} \left( 1 - \frac{1}{4^2 + 4 + 1} \right) = \frac{1}{2} \left( 1 - \frac{1}{21} \right) = \frac{1}{2} \cdot \frac{20}{21} = \frac{10}{21} \] Answer: \( \boxed{\frac{10}{21}} \)

Q2. Evaluate the sum of first 5 terms: \[ \sum_{k=1}^{5} \frac{k}{1 + k^2 + k^4} \]

Click to view solution

Using the formula:

\[ = \frac{1}{2} \left( 1 - \frac{1}{5^2 + 5 + 1} \right) = \frac{1}{2} \left( 1 - \frac{1}{31} \right) = \frac{1}{2} \cdot \frac{30}{31} = \frac{15}{31} \] Answer: \( \boxed{\frac{15}{31}} \)

Q3. If the sum of the series \[ \sum_{k=1}^{n} \frac{k}{1 + k^2 + k^4} \] is \( \frac{49}{50} \), find the value of \( n \).

Click to view solution

We know:

\[ \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right) = \frac{49}{50} \] Multiply both sides by 2: \[ 1 - \frac{1}{n^2 + n + 1} = \frac{98}{50} = \frac{49}{25} \] Now solve: \[ \frac{1}{n^2 + n + 1} = 1 - \frac{49}{25} = -\frac{24}{25} \] This is not possible — LHS is always positive. Let's double check. Wait: our assumption is wrong. Start again. From: \[ \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right) = \frac{49}{50} \Rightarrow 1 - \frac{1}{n^2 + n + 1} = \frac{98}{50} = \frac{49}{25} \] Again not possible. Let's do it correctly. Multiply both sides: \[ 1 - \frac{1}{n^2 + n + 1} = \frac{98}{100} = \frac{49}{50} \Rightarrow \frac{1}{n^2 + n + 1} = 1 - \frac{49}{50} = \frac{1}{50} \Rightarrow n^2 + n + 1 = 50 \Rightarrow n^2 + n - 49 = 0 \Rightarrow n = \frac{-1 \pm \sqrt{1 + 196}}{2} = \frac{-1 \pm \sqrt{197}}{2} \] No integer value. So, check if: \[ \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right) = \frac{49}{50} \Rightarrow 1 - \frac{1}{n^2 + n + 1} = \frac{98}{100} = \frac{49}{50} \Rightarrow \frac{1}{n^2 + n + 1} = \frac{1}{50} \Rightarrow n^2 + n + 1 = 50 \Rightarrow n^2 + n - 49 = 0 \] Again, discriminant is 1 + 4×49 = 197 ⇒ no integer \( n \). Therefore, there's no integer value of \( n \) for which the sum is exactly \( \frac{49}{50} \). Answer: No integer value of \( n \) satisfies the condition.

To Check More Sequence and Series Questions Click Here