Question

Adding consecutive odd numbers, find the perfect cube of 5.

Answer 1

Step-by-step explanation:

Answer 2

Answer:Si=1n+1 [(n + 1)n + (2i - 1)]    

Step-by-step explanation: (n + 1)3                  =  n3  + 3n2 + 3n + 1

= Si=1n [n (n - 1) + (2i - 1)] + 3n2 + 3n + 1      [by (5)]

=  n[n(n - 1)] + [1 + 3 + ... + (2n -1)] + 3n2 + 3n + 1

=  [n3  + 2n2 + 3n + 1] + [1 + 3 + ... + (2n -1)]          

= [n3  + 2n2 + 3n + 1] - (2n+1) +  [1 + 3 + ... + (2n -1)] +(2n+1)

= (n2 + 2n +1)n + S i=1n+1 (2i - 1) = (n + 1)(n + 1)n + S i=1n+1 (2i - 1)

= Si=1n+1 [(n + 1)n + (2i - 1)]