A useful integral identity

$(\int_{t}^{T}\sigma(t,u)du)^2-\int_{t}^{T}\sigma (t,u)\Sigma (t,u)du\equiv 0$ where $\Sigma (t,T)=\int_{t}^{T}\sigma(t,u)du$
This is the identity I found when I derive bond price under Q measure in the homework.

Power sum

sum(i) = n(n+1)/2
sum(i^2) = n(n+1)(2n+1)/6
sum(i^3) = n^4/4 + n^3/2 + n^2/2
...