Given that ((_r^n)) = (^nC_r), simplify ((^{2x + 1}_{3})) – ((^{2x – 1}_3)) – 2((^x_2))
Explanation
((^{2x + 1}_{3})) = (frac{(2x + 1)!}{3!(2x – 2)!} = frac{(2x + 1)2x(2x – 1)}{6})
(2x – 1) = (frac{(2x – 1)!}{3!(2x – 4)!} = frac{(2x – 1)(2x -2)(2x – 3)}{6}) and 2((^x_2)) = 2 [(frac{x!}{2!(x – 2)!})]
2[(frac{x(x – 1)}{2})] = x(x – 1)
Substitute; ((^{2x + 1}_3)) – ((^{2x + 1}_3)) – 2((^x_2))
(frac{(2x + 1) 2x(2x – 1)}{6} – frac{(2x – 1)(2x – 2)(2x – 3)}{6} – x(x – 1))
Multiplying through by 6 and simplifying to arrive at; 3(x^2 – 3x + 1)