Question

Solve the following expression : $\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}}$
A. Cannot be determined
B. $\left( {n - 1} \right)\left( {2n - 1} \right)$
C. None of these
D. $\left( {n + 1} \right)\left( {2n + 1} \right)$

Answer 1

The given problem revolves around the concepts of the equation of summation for the sequence containing squares, cubes, etc. So, we will first analyze the given expression with general formulae $\sum\limits_{i = 1}^n {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}$which is the squares sequence. Then, by substituting the given expression with the standard formula, the desired solution can be obtained.

Complete step by step answer:
Since, we have given the expression that
$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}}$
Since, it seems that after substituting the one by one values serially from one that is $k = 1$, we get
$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = {1^2} - {2^2} + {3^2} - {4^2} + .... - {\left( {2n} \right)^2} + {\left( {2n + 1} \right)^2}$
Separating the negative as well as positive terms in one bracket, we get
$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \left[ {{1^2} + {3^2} + .... + {{\left( {2n + 1} \right)}^2}} \right] - \left[ {{2^2} + {4^2} + .... + {{\left( {2n} \right)}^2}} \right]$
Combining both the brackets, we get

\Rightarrow \sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \sum\limits_{k = 1}^{2n + 1} {{{\left( {2n + 1} \right)}^2} - 8\sum\limits_{i = 1}^n {\left( {{1^2} + {2^2} + {3^2} + ... + {n^2}} \right)} } \\\ $$ … (i) Now, we know that the summation of any square of nth term is always, $\sum\limits_{i = 1}^n {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} $ So using this formula the required solution is; Let us solve one by one, first of all we will consider $$\sum\limits_{i = 1}^n {{n^2} = {{\left( {2n + 1} \right)}^2}} $$ term, we get Using the above mentioned formula substituting the respective term, we get $$\sum\limits_{i = 1}^n {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} $$ But, there is ‘$n$’ tends to ‘$2n + 1$that is$n \to 2n + 1$ As a result, we get $$\sum\limits_{i = 1}^n {{{\left( {2n + 1} \right)}^2} = \dfrac{{\left( {2n + 1} \right)\left( {2 + n + 1} \right)\left[ {2\left( {2n + 1} \right) + 1} \right]}}{6}} $$ Similarly, considering the second term$$\sum\limits_{k = 1}^{2n + 1} {{{\left( {2n + 1} \right)}^2} = } 2{\left( 2 \right)^2}\left[ {{1^2} + {2^2} + {3^2} + ... + {n^2}} \right]$$, we get $$\sum\limits_{k = 1}^{2n + 1} {{{\left( {2n + 1} \right)}^2} = } 2{\left( 2 \right)^2}\left[ {{1^2} + {2^2} + {3^2} + ... + {n^2}} \right] \\\ \Rightarrow \sum\limits_{k = 1}^{2n + 1} {{{\left( {2n + 1} \right)}^2} = 8} \sum\limits_{k = 1}^{2n + 1} {\left( {{1^2} + {2^2} + {3^2} + ... + {n^2}} \right)} \\\ $$ Solving the equation mathematically, we get $$\sum\limits_{k = 1}^{2n + 1} {{n^2} = 8} \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}$$ Now, the equation (i) becomes $$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \dfrac{{\left( {2n + 1} \right)\left( {2n + 1 + 1} \right)\left[ {2\left( {2n + 1} \right) + 1} \right] - 8n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \\\ \Rightarrow \sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \dfrac{{\left( {2n + 1} \right)}}{6}\left[ {\left( {2n + 2} \right)\left[ {4n + 2 + 1} \right] - 8n\left( {n + 1} \right)} \right] \\\ $$ Significantly, solving the equation, we get $$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \dfrac{{\left( {2n + 1} \right)}}{6}\left[ {2\left( {n + 1} \right)\left[ {4n + 3} \right] - 8n\left( {n + 1} \right)} \right]$$ Taking $n + 1$common from the equation, we get $$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \dfrac{{\left( {2n + 1} \right)\left( {n + 1} \right)}}{6}\left[ {8n + 6 - 8n} \right]$$ Solving the equation mathematically, we get $$\sum\limits_{k = 1}^{2n + 1} {{{\left( { - 1} \right)}^{k - 1}}{k^2}} = \left( {2n + 1} \right)\left( {n + 1} \right)$$ **Hence, option D is correct.** **Note:** One must be able to compare the given such expression/s of summation algebraically being asked to solve! Here, the expression is contains its square which is represented by $\sum\limits_{i = 1}^n {{n^2} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} $. Absolute expansion, substitution and solution is necessary, so as to be sure of our final answer.