Prove 1 + 5 + 9 + ... + (4n - 3) = n(2n - 1), `AA n in N`

Prove the following by the method of induction for all n in N : 1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = n/3 (2n -1) (2n +1)

Find the sum 1 xx 3 xx 5 + 3 xx 5 xx 7 + 5 xx 7 xx 9 + ……. + (2n - 1) ( 2n + 1) ( 2n + 3)

By method of induction prove that 1.3 + 2.5 + 3.7 +...+ n (2n + 1) = n/6 (n + 1) (4n + 5) for all n in N

The value of 2^n [1.3.5....(2n - 3)(2n - 1) is

Prove, by method of induction, for all n in N : 2 + 3.2 +4.2^2 +...+ (n + 1) (2^(n-1)) = n.2^n

Prove, by method of induction, for all n in N : 1^2 + 4^2 + 7^2 +...+ (3n - 2)^2 = n/2 (6n^2 - 3n - 1)

Prove, by method of induction, for all n in N : 8 + 17 + 26 +...+ (9n - 1) = n/2 (9n + 7)

Prove the following by the method of induction for all n in N : 1^2 + 2^2 + 3^2 + ... + n^2 = (n(n + 1) (2n + 1)) / 6 .

By method of induction, prove that 1.6 + 2.9 + 3.12 +... +to n terms = n(n +1 )(n+2) for all n in N .