Question

What is the sum of the series ${{1}^{3}}-{{2}^{3}}+{{3}^{3}}-...+{{9}^{3}}$ ?

Answer 1

Hint : To find the sum of the series, we will find the general term by separating the positive and the negative terms. To make the general term simpler we can find the general term of the simple summation series and subtract twice the general term of the negative terms.

Complete step-by-step answer :
The given series shows that the sum of the cube of the even terms up to $9$ is subtracted from the sum of the cube of odd terms up to $9$ .
Let us denote the sum of the series by $S$ .
Now, let us separate the odd and the even terms with their respective signs,
$S={{1}^{3}}+{{3}^{3}}+{{5}^{3}}+{{7}^{3}}+{{9}^{3}}-{{2}^{3}}-{{4}^{3}}-{{6}^{3}}-{{8}^{3}}$
Taking the negative sign common from the even numbers,
$\therefore S={{1}^{3}}+{{3}^{3}}+{{5}^{3}}+{{7}^{3}}+{{9}^{3}}-({{2}^{3}}+{{4}^{3}}+{{6}^{3}}+{{8}^{3}})$
Here, the general term for the odd terms is found as ${{(2n-1)}^{3}}$ , where $n=1,2,3,4,5$
However, simplifying this term becomes very complex.
Hence, here we will modify the series as follows,
$\therefore S={{1}^{3}}+{{2}^{3}}-2\left( {{2}^{3}} \right)+{{3}^{3}}+{{4}^{3}}-2\left( {{4}^{3}} \right)+{{5}^{3}}+{{6}^{3}}-2\left( {{6}^{3}} \right)+{{7}^{3}}+{{8}^{3}}-2\left( {{8}^{3}} \right)+{{9}^{3}}$
Now, separating the positive and the negative terms,
$\therefore S={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{6}^{3}}+{{7}^{3}}+{{8}^{3}}+{{9}^{3}}-2\left( {{2}^{3}} \right)-2\left( {{4}^{3}} \right)-2\left( {{6}^{3}} \right)-2\left( {{8}^{3}} \right)$
Taking the common part out of parenthesis of the negative terms,
$\therefore S={{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+{{6}^{3}}+{{7}^{3}}+{{8}^{3}}+{{9}^{3}}-2\left( {{2}^{3}}+{{4}^{3}}+{{6}^{3}}+{{8}^{3}} \right)$
Now, the general term for the natural numbers is $n$
And the general term for even terms is $2\;n$
For the given question, the general term for the sum of cubes of natural numbers is ${{n}^{3}}$ , where $n=1,2,3,...,9$
The general term for the cubes of even numbers is ${{\left( 2n \right)}^{3}}$ , where $n=1,2,3,4$
Now, applying the summation $n$ the form of a general term,
$\therefore S=\sum\limits_{n=1}^{9}{{{n}^{3}}}-2\left( \sum\limits_{n=1}^{4}{{{\left( 2n \right)}^{3}}} \right)$
$\therefore S=\sum\limits_{n=1}^{9}{{{n}^{3}}}-2\left( \sum\limits_{n=1}^{4}{8{{n}^{3}}} \right)$
Taking the constant part out of the summation,
$\therefore S=\sum\limits_{n=1}^{9}{{{n}^{3}}}-2\times 8\left( \sum\limits_{n=1}^{4}{{{n}^{3}}} \right)$
$\therefore S=\sum\limits_{n=1}^{9}{{{n}^{3}}}-16\left( \sum\limits_{n=1}^{4}{{{n}^{3}}} \right)$
Now, the summation for ${{n}^{3}}$ is given as
$\sum{{{n}^{3}}=\dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4}}$
Substituting in the equation,
$\therefore S={{\left[ \dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4} \right]}_{n=9}}-16{{\left[ \dfrac{{{n}^{2}}{{(n+1)}^{2}}}{4} \right]}_{n=4}}$
Substituting the value of $n$ ,
$\therefore S=\left[ \dfrac{{{(9)}^{2}}{{(9+1)}^{2}}}{4} \right]-16\left[ \dfrac{{{(4)}^{2}}{{(4+1)}^{2}}}{4} \right]$
$\therefore S=\left[ \dfrac{81\times 100}{4} \right]-16\left[ \dfrac{16\times 25}{4} \right]$
Removing the common factors from numerator and denominator,
$\therefore S=\left[ 81\times 25 \right]-16\left[ 4\times 25 \right]$
$\therefore S=\left[ 81\times 25 \right]-\left[ 64\times 25 \right]$
Taking the common factor out of parenthesis,
$\therefore S=25\left( 81-64 \right)$
$\therefore S=25\left( 17 \right)$
Hence, we get the final sum of the given series as
$\therefore S=425$
So, the correct answer is “425”.

Note : Here, we can simply find the cube of all the numbers individually and then add the cubes of all even numbers, and subtract the cubes of all odd numbers from the total. However, this method requires us to remember cubes of all the numbers.