Sum of cubes of the first n natural numbers - Formula derivation
Hello, Everyone. In this, we are going to derive the formula to find the sum of cubes of the first n natural numbers. Let S = 1 3 + 2 3 + 3 3 + .......... + (n-1) 3 + n 3 (m+1) 4 - m 4 = (m 4 + 4m 3 + 6m 2 + 4m + 1) - m 4 = 4m 3 + 6m 2 + 4m + 1 ------------ (i) put m = 1,2,3, ....... , (n-1), n in (i) Then, 2 4 - 1 4 = 4(1) 3 + 6(1) 2 + 4(1) + 1 ------------ (1) 3 4 - 2 4 = 4(2) 3 + 6(2) 2 + 4(2) + 1 ------------ (2) 4 4 - 3 4 = 4(3) 3 + 6(3) 2 + 4(3) + 1 ------------ (3) ................... n 4 - (n-1) 4 = 4(n-1) 3 + 6(n-1) 2 + 4(n-1) + 1 ------------ (n-1) (n+1) 4 - n 4 = 4(n) 3 + 6(n) 2 + 4(n) + 1 ------------ (n) Now, by adding equations (1),(2),(3), ...... ,(n-1) and (n), we get the following equation. (n+1) 4...