Noticing that each coefficient is the exact average of the two surrounding coefficients allows for a solution without calculus.

If we represent f=x+2x^2+3x^3+...+nx^n with [1,2,3,...,n,0,0,0...], then xf is [0,1,2,3,...,n-1,n,0,0...] and x²f is [0,0,1,2,3,...n-2,n-1,n,0,...].

Doing the numbers, x²f-2xf+f turns out to be [-1, 0, 0, ..., 0, -(n+1), n, 0...] or nx^(n+2)-(n+1)x^(n+1), from which we get the formula that was already found.