Question

Knowing the formula to the sum of the N integers a) what is the sum of the first N consecutive square integers, `Sigma_(k=1)^N k^2 = 1^2+2^2+ cdots + (N-1)^2+N^2`? b) Sum of the first N consecutive cube integers `Sigma_(k=1)^N k^3`?

Answer 1

For `S_k(n) = sum_{i =0}^n i^k`
`S_1(n) = (n(n+1))/2`
`S_2(n)=1/6 n (1 + n) (1 + 2 n)`
`S_3(n)=((n+1)^4-(n+1)-6S_2(n)-4S_1(n))/4`

Explanation:

We have

`sum_{i=0}^n i^3=sum_{i=0}^n (i+1)^3 - (n+1)^3`
`sum_{i=0}^n i^3=sum_{i=0}^n i^3 +3sum_{i=0}^n i^2+3sum_{i=0}^n i +sum_{i=0}^n 1- (n+1)^3`
`0 = 3sum_{i=0}^n i^2+3sum_{i=0}^n i +sum_{i=0}^n 1- (n+1)^3`

solving for `sum_{i=0}^n i^2`

`sum_{i=0}^n i^2=(n+1)^3/3-(n+1)/3-sum_{i=0}^n i`
but `sum_{i=0}^n i =((n+1)n)/2` so
`sum_{i=0}^n i^2=(n+1)^3/3-(n+1)/3-((n+1)n)/2`

`sum_{i=0}^n i^2=1/6 n (1 + n) (1 + 2 n)`

Using the same procedure for `sum_{i=0}^n i^3`

`sum_{i=0}^n i^4=sum_{i=0}^n (i+1)^4 - (n+1)^4`
`sum_{i=0}^n i^4=sum_{i=0}^n i^4 +4sum_{i=0}^n i^3+6sum_{i=0}^n i^2 + 4sum_{i=0}^n i +sum_{i=0}^n 1- (n+1)^4`

`0 = 4sum_{i=0}^n i^3+6sum_{i=0}^n i^2 + 4sum_{i=0}^n i +sum_{i=0}^n 1- (n+1)^4`
`0=4S_3(n) + 6S_2(n)+4S_1(n)+(n+1)-(n+1)^4`

Solving for `S_3(n)`

`S_3(n)=((n+1)^4-(n+1)-6S_2(n)-4S_1(n))/4`

Here `S_k(n) = sum_{i =0}^n i^k`