Question

The sum of the $n$ terms of the following series ${1^3} + {3^3} + {5^3} + {7^3} + ....$ is:
1) ${n^2}\left( {2{n^2} - 1} \right)$
2) ${n^3}\left( {n - 1} \right)$
3) ${n^3} + 8n + 4$
4) $2{n^4} + 3{n^2}$

Answer 1

Hint: We will first use the formula of sum of $n$ terms of ${1^3} + {2^3} + {3^3} + {4^3} + ....{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$. Find the sum of ${1^3} + {2^3} + {3^3} + {4^3} + ....{\left( {2n} \right)^3}$ and subtract the sum ${2^3} + {4^3} + {6^3} + {8^3} + ....{\left( {2n} \right)^3}$ to find the required answer.

Complete step by step answer:
We know a formula involving cube, that is the formula of the sum of cube of $n$ terms which is, ${1^3} + {2^3} + {3^3} + {4^3} + ....{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$.
Now, in the above formula, we have both odd and even terms. We need to subtract the sum of even terms from the known formula to get the desired answer.
Hence, we will first calculate the sum of cube of even terms.
${2^3} + {4^3} + {6^3} + {8^3} + ....{\left( {2n} \right)^3}$
After taking, ${2^3}$ common from the above series, we get,
${2^3}\left( {{1^3} + {2^3} + {3^3} + {4^3} + ....{n^3}} \right)$
We can solve the above expression using the formula, ${1^3} + {2^3} + {3^3} + {4^3} + ....{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$
Hence, the sum of ${2^3} + {4^3} + {6^3} + {8^3} + ....{\left( {2n} \right)^3}$ is ${2^3}\left( {\dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}} \right) = 2{n^2}{\left( {n + 1} \right)^2}$.
Also, sum of ${1^3} + {2^3} + {3^3} + {4^3} + ...{\left( {2n} \right)^3}$ is $\dfrac{{{{\left( {2n} \right)}^2}{{\left( {2n + 1} \right)}^2}}}{4} = {n^2}{\left( {2n + 1} \right)^2}$, where $2n$ is the total number of terms.
To find the sum of cube of odd numbers, the subtract the sum of even terms from ${1^3} + {2^3} + {3^3} + {4^3} + ...{\left( {2n} \right)^3}$.
${1^3} + {2^3} + {3^3} + {4^3} + ....{\left( {2n} \right)^3} - \left( {{2^3} + {4^3} + {6^3} + {8^3} + ....{{\left( {2n} \right)}^3}} \right) = {1^3} + {3^3} + {5^3} + {7^3} + ....{n^3}$
On substituting the values, ${1^3} + {2^3} + {3^3} + {4^3} + ....{\left( {2n} \right)^3} = {n^2}{\left( {2n + 1} \right)^2}$ and ${2^3} + {4^3} + {6^3} + {8^3} + .... = 2{n^2}{\left( {n + 1} \right)^2}$, we get,
$
  {1^3} + {3^3} + {5^3} + {7^3} + .... = {n^2}{\left( {2n + 1} \right)^2} - 2{n^2}{\left( {n + 1} \right)^2} \\
  {1^3} + {3^3} + {5^3} + {7^3} + .... = \left( {4{n^4} + 4{n^3} + {n^2}} \right) - 2\left( {{n^4} + 2{n^3} + {n^2}} \right) \\
  {1^3} + {3^3} + {5^3} + {7^3} + .... = {n^2}\left( {2{n^2} - 1} \right) \\
 $
 Therefore, the sum of ${1^3} + {3^3} + {5^3} + {7^3} + .... = {n^2}\left( {2{n^2} - 1} \right)$
Hence, option A is the correct answer.

Note:- We find the sum of cubes of $2n$ number, as after subtracting half of the terms, we will be left with $n$ terms, as required in the question. It is known that the sum of cubes of first $n$ natural numbers is ${1^3} + {2^3} + {3^3} + {4^3} + ....{n^3} = \dfrac{{{n^2}{{\left( {n + 1} \right)}^2}}}{4}$.