How could I compute the formula for ∑ i^2 (i=1 to n) ???

Question

I have to compute it using the telescoping sum; I know what the formula is but how would I compute it???

Answer 1

One possible method of deriving the formula is to compute the sum of (k+1)³ (sic) in two different ways:

[k=1, n]∑(k+1)³
= [k=2, n+1]∑k³
= (n+1)³ - 1 + [k=1, n]∑k³

And:

[k=1, n]∑(k+1)³
= [k=1, n]∑(k³ + 3k² + 3k + 1)
= [k=1, n]∑k³ + 3[k=1, n]∑k² + 3[k=1, n]∑k + [k=1, n]∑1

So we have that:

(n+1)³ - 1 + [k=1, n]∑k³ = [k=1, n]∑k³ + 3[k=1, n]∑k² + 3[k=1, n]∑k + [k=1, n]∑1

And subtracting [k=1, n]∑k³ from both sides yields:

(n+1)³ - 1 = 3[k=1, n]∑k² + 3[k=1, n]∑k + [k=1, n]∑1

And since the formulae for [k=1, n]∑1 and [k=1, n]∑k are easy to deduce, we may substitute them to find:

(n+1)³ - 1 = 3[k=1, n]∑k² + 3n(n+1)/2 + n

If we denote [k=1, n]∑k² by S, we now have an equation that we can solve for S:

(n+1)³ - 1 = 3S + 3n(n+1)/2 + n
n³ + 3n² + 3n + 1 - 1 = 3S + 3n(n+1)/2 + n
2n³ + 6n² + 6n = 6S + 3n(n+1) + 2n
2n³ + 6n² + 6n = 6S + 3n² + 3n + 2n
2n³ + 3n² + n = 6S
n(n+1)(2n+1) = 6S
S = n(n+1)(2n+1)/6
[k=1, n]∑k² = n(n+1)(2n+1)/6

And we are done. Easier than falling off a logarithm.