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If $1 + 2 + 3 + ..... + n = k$, then ${1^3} + {2^3} + {3^3} + .......... + {n^3}$ is equal to
$
  {\text{A}}{\text{. }}{k^2} \\
  {\text{B}}{\text{. }}{k^3} \\
  {\text{C}}{\text{. }}\dfrac{{k(k + 1)}}{2} \\
  {\text{D}}{\text{. }}{\left( {k + 1} \right)^3} \\
 $

seo-qna
Last updated date: 25th Jul 2024
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Answer
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64.2k+ views
Hint- Here, we will proceed by using the formula for the sum of first \[n\] natural numbers.

Given, $1 + 2 + 3 + ..... + n = k$
Since, we know that the sum of first \[n\] natural numbers i.e., \[1 + 2 + 3 + ..... + n = \dfrac{{n\left( {n + 1} \right)}}{2}\]
\[ \Rightarrow k = \dfrac{{n\left( {n + 1} \right)}}{2}{\text{ }} \to {\text{(1)}}\]
Also, the sum of cubes of first \[n\] natural numbers i.e., \[{1^3} + {2^3} + {3^3} + .......... + {n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2}\]
Using equation (1), we have
\[{1^3} + {2^3} + {3^3} + .......... + {n^3} = {\left[ {\dfrac{{n\left( {n + 1} \right)}}{2}} \right]^2} = {k^2}\]
Therefore, the sum of cubes of first \[n\] natural numbers i.e., \[{1^3} + {2^3} + {3^3} + .......... + {n^3} = {k^2}\]
Therefore, option A is correct.

Note- In these type of problems, we will simply be using some general formulas like sum of first \[n\] natural numbers, sum of squares of first \[n\] natural numbers and sum of cubes of first \[n\] natural numbers which will redirect us to the final answer.