Answer:
Explanation:
1) The sum of a series of the cubes of numbers 'i' is given as follows;
Sₙ = n/2 × (1st term + Last term)
∴ Sₙ = n/2 × (1 + n) = n·(n + 1)/2
It can be shown that the sum of a series of the cubes of numbers 'i²' is given as follows;
Sₙ = (n·(n + 1)·(2·n + 1))/6
It can be also be shown that the sum of a series of the cubes of numbers 'i³' is given as follows;
Sₙ = (n·(n + 1)/2)²
Therefore, we have;
.