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avatar+17 

\(1^3+2^3+3^3+...+n^3=(\frac{n(n+1)}{2})^2\)

 Dec 14, 2022
 #1
avatar+24 
-1

The sum of the cubes of the first n natural numbers is given by the formula: ((n(n+1)/2)^2 = (n^2(n+1)^2)/4 Therefore, the sum of the cubes of the first n natural numbers is ((n^2(n+1)^2)/4).

 Dec 14, 2022
 #2
avatar+33337 
+1

Have a look at https://mathschallenge.net/library/number/sum_of_cubes 

 Dec 14, 2022

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