If $A_{1}, A_{2}, A_{3}, A_{4}$ are any: four consecutive
binomial coefficients in the expansion of
$(1 + x)^{n}, then \frac{A_{1}}{A_{1} + A_{2}} + \frac{A_{3}}{A_{3} + A_{4}} - \frac{2 A_{2}}{A_{2} + A_{3}} =$

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$n - 2$

$1$

$- 1$

$0$

answer is D.

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Detailed Solution

$A_{1} = (\begin{matrix} n \\ r - 1 \end{matrix}), A_{2} = (\begin{array}{l} n \\ r \end{array}), A_{3} = (\begin{matrix} n \\ r + 1 \end{matrix}), A_{4} = (\begin{matrix} n \\ r + 2 \end{matrix})$

$\begin{aligned} \frac{A_{1}}{A_{1} + A_{2}} = \frac{1}{1 + \frac{A_{2}}{A_{1}}} = \frac{1}{1 + \{(\begin{matrix} n \\ r \end{matrix}) / (\begin{matrix} n \\ r - 1 \end{matrix})\}} \\ = \frac{1}{1 + \frac{n - r + 1}{r}} = \frac{r}{n + 1}, \frac{A_{2}}{A_{2} + A_{3}} = \frac{r + 1}{n + 1} \\ \frac{A_{3}}{A_{3} + A_{4}} = \frac{r + 2}{n + 1} \\ ∴ \frac{A_{1}}{A_{1} + A_{2}} + \frac{A_{3}}{A_{3} + A_{4}} - \frac{2 A_{2}}{A_{2} + A_{3}} = \frac{r}{n + 1} + \frac{r + 2}{n + 1} - \frac{2 (r + 1)}{n + 1} = 0 \end{aligned}$