Given that (\(_r^n\)) = \(^nC_r\), simplify…

Question 1 Report

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

Answer

Answer Details

Write each term using \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\).

\[\binom{2x+1}{3}=\frac{(2x+1)(2x)(2x-1)}{6}=\frac{8x^3-2x}{6},\] \[\binom{2x-1}{3}=\frac{(2x-1)(2x-2)(2x-3)}{6}=\frac{8x^3-24x^2+22x-6}{6}.\]

Their difference:

\[\binom{2x+1}{3}-\binom{2x-1}{3}=\frac{(8x^3-2x)-(8x^3-24x^2+22x-6)}{6}=\frac{24x^2-24x+6}{6}=4x^2-4x+1.\]

And \(2\binom{x}{2}=2\cdot\dfrac{x(x-1)}{2}=x^2-x\). Therefore

\[\binom{2x+1}{3}-\binom{2x-1}{3}-2\binom{x}{2}=(4x^2-4x+1)-(x^2-x)=3x^2-3x+1.\]

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