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# If $1 + 2 + 3 + ..... + n = k$, then ${1^3} + {2^3} + {3^3} + .......... + {n^3}$ is equal to${\text{A}}{\text{. }}{k^2} \\ {\text{B}}{\text{. }}{k^3} \\ {\text{C}}{\text{. }}\dfrac{{k(k + 1)}}{2} \\ {\text{D}}{\text{. }}{\left( {k + 1} \right)^3} \\$ Verified
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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.
Last updated date: 24th Sep 2023
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