Can anyone check if I correctly derived the K^2 summation formula?

Question

I have no idea how to upload a picture here. So here is the link to how I derived the formula. I would greatly appreciate it if you could give me any feedback on my work. Thanks in advance.

Link: https://imgur.com/ODzIXsg

Answer 1

Not sure if a "correct" derivation exists, yet the traditional on is started:

a³ - b³ = (a - b)(a² + ab + b²)

Then)

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

= 1(n² + n² - n + n² - 2n + 1

= 3n² - 3n + 1

and

(n -1)³ - (n - 2)³ = 3(n-1)² - 3(n - 1) + +1

(n - 2)³ - (n - 3)³ = 3(n - 2)³ - 3(n - 2) + 1

continues to

2³ - 1³ = 3(2²) - 3(2) + 1

1³ - 0³ = 3(1²) 3(1) + 1

All left side terms add to n³ (Leave it to you to verify)

All the right side add to

3 ∑ⁿ_n=1 n² + 3∑ⁿ_n=1 n + n (Leave it to you to verify)

n³ = 3 ∑n² + 3 ∑ n + n

Now 3 ∑ n = 3(n)(n + 1) / 2

n³ = 3 ∑ n² + 3(n)(n +1) / 2 + n

Isolating ∑ n²

∑ n² = (1/3)[n³ + 3(n)(n + 1) / 2 -n]

Factoring out an "n" on the right side

∑ n² = (n/3)[n² + 3(n + 1) /2 - 1]

∑ n² = (n/6)[2n² + 3n + 1]

∑ n² = (n/6 )(2n +1)(n + 1)

∑ n² = n(n + 1)(2n + 1) / 6

QED