#8 Proof by induction Σ k^2= n(n+1)(2n+1)/6 discrete principle induccion matematicas

For more cool math videos visit my site at http://mathgotserved.com or http://youtube.com/mathsgotserved 5^n-2^n is divisible by 3, n^3+2n is divisible by 3, 8^n-1 is divisible by 7, 1/(1X2)+1/(2X3)+...+1/(n(n+1))=n/(n+1), n^2 less than2n, 8+2x8+3x8+..+8n=, 1+3+5+7+...+(2n-1)=n^2 for all n in N, sum k^2= 1/6n(n+1)(2n+1), 9^n-2^n id divisible by 7, 1+5+9+13+..+(4n-3)=½n(4n-2), n Σ k =n(n+1)/2 , 1^3+2^3+3^3+..+n^3= (n(n+1)/2)^2 n^2(n+1)^2/4, 10^1+10^2+10^3+ +10^n = 10/9(10^n-1), derivative of x^-1 is ( (-1)^n n !)/ x^(n+1), 2^n greater or equal 1+n, 3^n less thatn (n+1)!, 8^n-1 divisible by 7