Algebra Question

Question

If P(n): ∑^k_i=1 i² (i+1) = 1/12 k (k+1)(k+2)(3k+1)

Prove P(k+1) is true. With steps.

Answer 1

P(k)= ∑^k_i=1 i² (i+1) = 1/12 k (k+1)(k+2)(3k+1)

P(k+1)= ∑^k+1_i=1 i² (i+1) = P(k)+ (k+1)² (k+2)

=1/12 k (k+1)(k+2)(3k+1) + (k+1)² (k+2) =|Simplify it and you have to obtain|

P(k+1) = (1/12) (k+1)(k+2)(k+3)(3k+4)