If... $x^{x + 1} C_{3} -^{x - 1} C_{3} = 16 x = ?$ $color(white)x^(x + 1)C_3 - ^(x - 1)C_3 = 16x = ?$

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Answer 1 · 2017-11-22T08:58:44

$x = 5$ $x=5$

$C_{3}^{x + 1} = \frac{(x + 1) x (x - 1)}{1.2 .3}$ $C_3^(x+1)=((x+1)x(x-1))/(1.2.3)$

and $C_{3}^{x - 1} = \frac{(x - 1) (x - 2) (x - 3)}{1.2 .3}$ $C_3^(x-1)=((x-1)(x-2)(x-3))/(1.2.3)$

Therefore $C_{3}^{x + 1} - C_{3}^{x - 1}$ $C_3^(x+1)-C_3^(x-1)$

= $\frac{(x + 1) x (x - 1)}{1.2 .3} - \frac{(x - 1) (x - 2) (x - 3)}{1.2 .3}$ $((x+1)x(x-1))/(1.2.3)-((x-1)(x-2)(x-3))/(1.2.3)$

= $\frac{x - 1}{6} {x (x + 1) - (x - 2) (x - 3)}$ $(x-1)/6{x(x+1)-(x-2)(x-3)}$

= $\frac{x - 1}{6} {x^{2} + x - (x^{2} - 5 x + 6)}$ $(x-1)/6{x^2+x-(x^2-5x+6)}$

= $\frac{x - 1}{6} (6 x - 6)$ $(x-1)/6(6x-6)$

= ${(x - 1)}^{2}$ $(x-1)^2$

Hence $C_{3}^{x + 1} - C_{3}^{x - 1} = 16 x$ $C_3^(x+1)-C_3^(x-1)=16x$ is equivalent to

${(x - 1)}^{2} = 16$ $(x-1)^2=16$

or $x - 1 = \pm 4$ $x-1=+-4$

i.e. $x = 5$ $x=5$ or $- 3$ $-3$

and if we consider only as a positive integer $x = 5$ $x=5$

If... color(white)x^(x + 1)C_3 - ^(x - 1)C_3 = 16x = ?xx+1C3−x−1C3=16x=?color(white)x^(x + 1)C_3 - ^(x - 1)C_3 = 16x = ?