(iii) Here S_n=1³+2³+3³+ …+n³
We consider the identity,

Putting k= 1, 2, 3, …, n, we get

Adding both sides, we get
(n+1)⁴-1⁴=4∑ⁿ_k=1 k³ +6∑ⁿ_k=1 n² +4∑ⁿ_k=1 n+n…(iii)
From parts (i) and (ii), we know that
∑ⁿ_k=1 k=½n(n+1) and ∑ⁿ_k=1 k² =⅙n(2n+1)(n+1)
Putting these values in equation (iii), we obtain

It is wonderful that ∑ⁿ_k=1 k³ =(∑ⁿ_k=1 k)².

For n= 3, 4, check:

📌 Example 1. Evaluate ∑⁶_i=1 (i³–i² ) .
Use Summation Properties first then apply the formulae (with n=6).

📌 Example 2. Evaluate

✍ Solution:
Use summation notation and summation formula, the numerator becomes

📌 Example 3. Find the value of the sum ∑²⁰⁰_k=1 (2k³-6k²+3) .

Let’s use the same approach as in the previous example. First, We’ll use the properties to split this into individual sums, then factor out the coefficients. After that, we’ll use the formulas above to evaluate it.

The numbers in this example were horribly ugly, but we were able to evaluate the sum without having to actually calculate all 200 terms, then add them all up. In 5 small lines, We were able to add 200 numbers.