Given that (\(_r^n\)) = \(^nC_r\), simplify (\(^{2x + 1}_{3}\)) - (\(^{2x - 1}_3\)) - 2(\(^x_2\))
FURTHER MATHEMATICS
WAEC 2019
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\(x^2 - 3x + 1\)
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