Question

For any odd integer $n\ge 1,{{n}^{3}}-{{\left( n-1 \right)}^{3}}+\ldots \ldots +{{\left( -1 \right)}^{n-1}}{{1}^{3}}=\ldots $ ?

Answer 1

Hint: For solving this question you should know about the expansion and solving of the expressions. In this problem we will first expand the given expression and then put the odd values for this and get the expressions. And then we will put the even values and get another expression for this.

Complete step by step answer:
According to our question it is asked to us that for any odd integer $n\ge 1,{{n}^{3}}-{{\left( n-1 \right)}^{3}}+\ldots \ldots +{{\left( -1 \right)}^{n-1}}{{1}^{3}}=\ldots $, we have to find the value. So, as we know that if we find the value of any expression then we always expand this for three to four powered digits and then we put our integers as required for this either even or odd.
Since $n$ is an odd integer ${{\left( -1 \right)}^{n}}=1$,
$\begin{align}
  & {{n}^{3}}-{{\left( n-1 \right)}^{3}}+{{\left( n-2 \right)}^{3}}-{{\left( n-3 \right)}^{3}}+\ldots \ldots +{{1}^{3}} \\
 & ={{n}^{3}}-{{\left( n-1 \right)}^{3}}+{{\left( n-2 \right)}^{3}}+\ldots \ldots +{{1}^{3}}-2\left[ {{\left( n-1 \right)}^{3}}+{{\left( n-2 \right)}^{3}}+\ldots \ldots {{2}^{3}} \right] \\
\end{align}$
And if we solve this equation, then,
$={{n}^{3}}+{{\left( n-1 \right)}^{3}}+{{\left( n-2 \right)}^{3}}+\ldots \ldots {{1}^{3}}-2\times {{2}^{3}}.\left[ {{\left( \dfrac{\left( n-1 \right)}{2} \right)}^{3}}+{{\left( \dfrac{\left( n-3 \right)}{2} \right)}^{3}}+\ldots \ldots {{1}^{3}} \right]$
So, this will be the expression for $n$ and here $n$ should be an odd integer.
And if $n-1,n-3$ are even integers, then,
$=\dfrac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}-16\dfrac{{{\left( n-1 \right)}^{2}}.{{\left( n+1 \right)}^{2}}}{16\times 4}$
And if we solve this, then we find,
$\begin{align}
  & =\dfrac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}-\dfrac{{{\left( n-1 \right)}^{2}}.{{\left( n+1 \right)}^{2}}}{4} \\
 & \Rightarrow \dfrac{{{n}^{2}}{{\left( n+1 \right)}^{2}}-{{\left( n-1 \right)}^{2}}.{{\left( n+1 \right)}^{2}}}{4} \\
 & \Rightarrow \dfrac{{{\left( n+1 \right)}^{2}}\left( {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right)}{4} \\
\end{align}$
$=\dfrac{1}{4}{{\left( n+1 \right)}^{2}}\left[ {{n}^{2}}-{{\left( n-1 \right)}^{2}} \right]$
On further simplifications, we get,
$\begin{align}
  & =\dfrac{1}{4}{{\left( n+1 \right)}^{2}}\left[ {{n}^{2}}-{{n}^{2}}+2n-1 \right] \\
 & =\dfrac{1}{4}{{\left( n+1 \right)}^{2}}\left[ 2n-1 \right] \\
\end{align}$
So, the final solution for this expression is $\dfrac{1}{4}{{\left( n+1 \right)}^{2}}\left[ 2n-1 \right]$.

Note: While solving this type of questions you have to be careful about the even and the odd values of the expressions because if we put another value then too, we can solve the expression, but then the answer will be different from the original answer and it will be incorrect.