M2 maths

A)when n = 1 , L.H.S. = 2^2 = 4 = R.H.S. = 2(1)(1+1)(2+1)/3 = 4
Assume when n = k the statement is true :
2^2+4^2+6^2+...+(2k)^2 = 2k(k+1)(2k+1)/3
when n = k+1 :
2^2+4^2+6^2+...+[2(k+1)]^2 = 2k(k+1)(2k+1)/3 + [2(k+1)]^2
R.H.S. = [2k(k+1)(2k+1) + 12(k+1)^2 ] / 3
= (k+1)[2k(2k+1) + 12(k+1)] / 3
= (k+1)(4k^2 + 2k + 12k + 12) / 3
= (k+1)(4k^2 + 14k + 12) / 3
= 2(k+1)(2k^2 + 7k + 6) / 3
= 2(k+1)(k+2)(2k+3) / 3
= 2(k+1)(k+2)[2(k+1) + 1] / 3
By mathematical induction it is true for all + integers.
B)1) 1^2+2^2+3^2+...+n^2
= [2^2 + 4^2 + 6^2 +...+ (2n)^2] / 4
= [2n(n+1)(2n+1)/3] / 4
= n(n+1)(2n+1) / 6
B2) 1^2+2^2+3^2+...+(2n)^2
By B1) :
= 2n(2n+1)(2(2n)+1) / 6
= n(2n+1)(4n+1) / 3
b3) (n+1)^2+(n+2)^2+...+(2n)^2
= (1^2+2^2+3^2+...+(2n)^2) - (1^2+2^2+3^2+...+n^2)
= n(2n+1)(4n+1)/3 - n(n+1)(2n+1)/6
= n(2n+1)(8n + 2 - n - 1) / 6
= n(2n+1)(7n+1) / 6