Proof for Sum of squares of first n natural,even,odd numbers - pending
We want to prove: ∑ k = 1 n k 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} Method1: Polynomial Assumption and Coefficient Matching The idea is to assume that the sum of squares is a cubic polynomial in n n : ∑ k = 1 n k 2 = A n 3 + B n 2 + C n + D \sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn + D But since ∑ k = 1 0 k 2 = 0 \sum_{k=1}^0 k^2 = 0 , we know that when n = 0 n = 0 , the sum is 0 ⇒ D = 0 D = 0 . So we assume: ∑ k = 1 n k 2 = A n 3 + B n 2 + C n \sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn Now compute the sum for small values of n n : n = 1 n = 1 : 1 2 = A ( 1 ) 3 + B ( 1 ) 2 + C ( 1 ) = A + B + C ⇒ A + B + C = 1 (1) 1^2 = A(1)^3 + B(1)^2 + C(1) = A + B + C \Rightarrow A + B + C = 1 \tag{1} n = 2 n = 2 : 1 2 + 2 2 = 1 + 4 = 5 ⇒ 8 A + 4 B + 2 C = 5 (2) 1^2 + 2^2 = 1 + 4 = 5 \Rightarrow 8A + 4B + 2C = 5 \tag{2} n = 3 n = 3 : 1 2 + 2 2 + 3 2 = 1 + 4 + 9 = 14 ⇒ 27 A + 9 B + 3 C = 14 (3) 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \Rightarrow 2...