Question

The average of cubes of natural numbers from 1 to 8

Answer 1

Sum ={n(n+1)/2}^2
(8*9/2)^2
36*36 1296
Average = 1296/8 =162

Answer 2

Answer:

Average = [1^3 + 2^3 + 3^3 + ______+8^3] / 8

              = {[8(8+1)/2]^2}/8

              = 1296/8

              = 162

Step-by-step explanation:

formula for sum of cubes of first n natural nos =  [n(n/2)]^2