<s>[INST] How to simplify this equation: $1\cdot N + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \dots + N \cdot1$? $$1 (N) + 2(N-1) + 3(N -2) + \cdots + i(N - i + 1) + \cdots + N (1)  $$
I need to write it in simplest form?
here 1(N) means 1 multiply by N 
 [/INST] A more complex but fun alternative solution using Finite Calculus:
$$
  \begin{align}
  　\sum_{x=1}^N x \left[(N+1)-x\right]
    & = \sum_{1}^{N+1} \left[ Nx - x(x-1)\right] \delta x
      = \sum_{1}^{N+1} \left( Nx^{(1)} - x^{(2)} \right) \delta x \\
    & = \left[ \frac{N}{2}x^{(2)} - \frac{1}{3} x^{(3)} \right]_{1}^{N+1}
      = \left[ \frac{N}{2}x(x-1) - \frac{1}{3} x(x-1)(x-2) \right]_{1}^{N+1}\\
    & = \left[ \frac{N^2(N+1)}{2} - 0 \right] - \left[ \frac{1}{3}(N+1)N(N-1) - 0 \right] \\
    & = \frac{N(N+1)(N+2)}{6}.
  \end{align}
$$
</s>