How to simplify this equation: $1\cdot N + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \dots + N \cdot1$?

Question

$$1 (N) + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \cdots + N (1) $$

I need to write it in simplest form?

here 1(N) means 1 multiply by N

Answer 1

$$ \sum_{k=1}^Nk(N-k+1)=\sum_{k=1}^N kN-\sum_{k=1}^Nk^2+\sum_{k=1}^Nk=N\,\frac{N(N+1)}2-\frac{N(N+1)(2N+1)}6+\frac{N(N+1)}2=\frac{N(N+1)}2\left(N-\frac{2N+1}3+1\right)=\frac{N(N+1)(N+2)}6. $$

Answer 2

Your sum can be written as

$$\sum_{k=1}^Nk(N-k+1) = N\sum_{k=1}^{N}k - \sum_{k=1}^N{k^2} + \sum_{k=1}^N{k}$$

$$ = \frac{N(N+1)^2}{2} - \frac{N(N+1)(2N+1)}{6}$$

Answer 3

A more complex but fun alternative solution using Finite Calculus:

$$ \begin{align} 　\sum_{x=1}^N x \left[(N+1)-x\right] & = \sum_{1}^{N+1} \left[ Nx - x(x-1)\right] \delta x = \sum_{1}^{N+1} \left( Nx^{(1)} - x^{(2)} \right) \delta x \\ & = \left[ \frac{N}{2}x^{(2)} - \frac{1}{3} x^{(3)} \right]_{1}^{N+1} = \left[ \frac{N}{2}x(x-1) - \frac{1}{3} x(x-1)(x-2) \right]_{1}^{N+1}\\ & = \left[ \frac{N^2(N+1)}{2} - 0 \right] - \left[ \frac{1}{3}(N+1)N(N-1) - 0 \right] \\ & = \frac{N(N+1)(N+2)}{6}. \end{align} $$