Solved: (2n+2)! Factorial problem - Experts Exchange The original form of this before putting it into An+1 An form was: summation of (2n)! (n!)^2 So of course the problem I wrote is in this form About the only thing about this problem I understand is that: (2n+2)! = (1)(2)(3) (2n)(2n+1)(2n +2) So in ths final answer, the numerator only consists of the last two terms of that sequence The (2n
How do you factor 2n^2 + 3n - 9? - Socratic 2n^2+ 3n -9 We can Split the Middle Term of this expression to factorise it In this technique, if we have to factorise an expression like an^2 + bn + c, we need to think of 2 numbers such that: N_1*N_2 = a*c = 2 xx (-9) = -18 and N_1 +N_2 = b = 3 After trying out a few numbers we get N_1 = 6 and N_2 =-3 6 xx (-3) = -18 and 6+(-3)= 3 2n^2+ 3n -9 = 2n^2+ 6n - 3n -9 = 2n(n +3) -3(n+3) =color