Explanation

(c)
Let $a_{1},a_{2},a_{3},a_{4}$ be respectively the coefficients of $(r + 1)$th, $(r + 2)$th, $(r + 3)$th and $(r + 4)$th terms in the expansion of ${(1 + x)}^{n}\$. Then ,
$a_{1} = \ ^{n}C_{r},\ a_{2} = \ ^{n}C_{r + 1},\ a_{3} = \ ^{n}C_{r + 2},\ a_{4} = \ ^{n}C_{r + 3}$
Now, $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{\ ^{n}C_{r}}{\ ^{n}C_{r} + \ ^{n}C_{r + 1}} + \frac{\ ^{n}C_{r + 2}}{\ ^{n}C_{r + 2} + \ ^{n}C_{r + 3}}$
$= \frac{\ ^{n}C_{r}}{\ ^{n + 1}C_{r + 1}} + \frac{\ ^{n}C_{r + 2}}{\ ^{n + 1}C_{r + 3}}\ \ \ (\because\ ^{n}C_{r} + \ ^{n}C_{r + 1} = \ ^{n + 1}C_{r + 1})$
$= \frac{\ ^{n}C_{r}}{\frac{n + 1}{r + 1}\ ^{n}C_{r}} + \frac{\ ^{n}C_{r + 2}}{\frac{n + 1}{r + 3}\ ^{n}C_{r + 2}}\ \ \ \left( \because\ ^{n}C_{r} = \frac{n}{r}\ ^{n - 1}C_{r - 2} \right)$
$= \frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2(r + 2)}{n + 1}$
$= 2\frac{\ ^{n}C_{r + 1}}{\ ^{n + 1}C_{r + 2}} = 2\frac{\ ^{n}C_{r + 1}}{\ ^{n}C_{r + 1} + \ ^{n}C_{r + 2}}$
$= \frac{2a_{2}}{a_{2} + a_{3}}$