Derivation of 1×2+2×3+3×4+....…+n(n+1) = [n(n+1)(n+2)]/3

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Derivation for r=∆/s Derivation of r = delta/s

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Sum of the squares of the first n natural numbers - Formula derivation

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