Question

The sum of \[\sum\limits_{k=1}^{100}{\dfrac{k}{{{k}^{4}}+{{k}^{2}}+1}}\] is equal to
A) \[\dfrac{4950}{10101}\]
B) \[\dfrac{5050}{10101}\]
C) \[\dfrac{5151}{10101}\]
D) None of these

Answer 1

Hint: In order to solve this question first we will use the property of sigma expansion and find the sum by using polynomial identities and the polynomial formula to simplify the Expression:
\[{{x}^{4}}+{{x}^{2}}+1=[({{x}^{2}}+1)+x][({{x}^{2}}+1)x]\]
\[={{({{x}^{2}}+1)}^{2}}{{(x)}^{2}}\]
\[={{x}^{4}}+1+2{{x}^{2}}{{x}^{2}}\]
\[={{x}^{4}}+{{x}^{2}}+1\]
Sigma expansion shows the sum of the series when put the limit of sigma in the expression. Summation of series means we add all the terms of the series that expands with the help of sigma expansion so we can identify the type of series and use the proper formula.

Complete step by step answer:
By using the formula of the polynomial we get,
\[{{k}^{4}}+{{k}^{2}}+1=[({{k}^{2}}+1)+k][({{k}^{2}}+1)k]\]
Simplify the expression using algebraic identity
\[={{({{k}^{2}}+1)}^{2}}{{k}^{2}}\]
Rewrite the expression after open the square
\[={{k}^{4}}+1+2{{k}^{2}}{{k}^{2}}\]
Rewrite the expression after simplification
\[={{k}^{4}}+{{k}^{2}}+1\]
We can also write the expression in the form
\[{{k}^{2}}+k+1{{k}^{2}}+k-1=2k\ \ldots \ldots \left( 1 \right)\]
Now, from the given expression
\[\Rightarrow \sum\limits_{k=1}^{100}{\dfrac{k}{{{k}^{4}}+{{k}^{2}}+1}}\]
Now, Substitute the value of \[{{k}^{4}}+{{k}^{2}}+1\] in the given expression
\[\Rightarrow \sum\limits_{k=1}^{100}{\dfrac{k}{({{k}^{2}}+1+k)({{k}^{2}}+1k)}}\]
Rewrite the expression after multiplication and division by \[2\]
\[\Rightarrow \dfrac{1}{2}\sum\limits_{k=1}^{100}{\dfrac{2k}{({{k}^{2}}+1+k)({{k}^{2}}k1)}}\]
Substitute the value of \[2k\] from the equation (1) in the expression
\[\Rightarrow \dfrac{1}{2}\sum\limits_{k=1}^{100}{\dfrac{({{k}^{2}}+k+1)({{k}^{2}}k+1)}{({{k}^{2}}+k+1)({{k}^{2}}k+1)}}\]
Simplify the expression by dividing the numerator by denominator
\[\Rightarrow \dfrac{1}{2}\sum\limits_{k=1}^{100}{\left\{ \dfrac{({{k}^{2}}+k+1)}{({{k}^{2}}+k+1)({{k}^{2}}k+1)}-\dfrac{({{k}^{2}}k+1)}{({{k}^{2}}+k+1)({{k}^{2}}k+1)} \right\}}\]
Rewrite the expression after simplification
\[\Rightarrow \dfrac{1}{2}\sum\limits_{k=1}^{100}{\left[ \dfrac{1}{{{k}^{2}}k+1}\dfrac{1}{{{k}^{2}}+k+1} \right]}\]
Put the value of limit and use the sigma expansion
\[\Rightarrow \dfrac{1}{2}\left[ \dfrac{1}{1}\dfrac{1}{3}+\dfrac{1}{3}+.........+\dfrac{1}{10101} \right]\]
Rewrite the expression after simplification
\[\Rightarrow \dfrac{1}{2}\left[ 1\dfrac{1}{10101} \right]\]
Again rewrite the expression after simplification
\[\Rightarrow \dfrac{1}{2}\left[ \dfrac{10101-1}{10101} \right]\]
Rewrite the expression after simplification
\[\Rightarrow \dfrac{1}{2}\left[ \dfrac{10100}{10101} \right]\]
Rewrite the expression after simplification
\[\therefore \left[ \dfrac{5050}{10101} \right]\]

Hence, option (B) is the correct answer.

Note:
Sigma expansion shows the sum of the series when put the limit of sigma in the expansion as
\[\sum\limits_{k=1}^{n}{k}=1+2+3\ \ldots \ldots +n\]
\[1+2+3\ \ldots \ldots +n=\dfrac{n\left( n+1 \right)}{2}\]
 We can also solve this problem by direct expansion by using the polynomial concept
\[{{x}^{4}}+{{x}^{2}}+1=[({{x}^{2}}+1)+x[({{x}^{2}}+1)x]\]
This is the form of polynomial making in the perfect square type.