If $a_{1}, a_{2}, a_{3}, a_{4}$ are the coeffecients of any four consecutive term in the expansion of ${(1 + x)}^{n}$ , then $\frac{a_{1}}{a_{1} {+ a}_{2}} + \frac{a_{3}}{a_{3} {+ a}_{4}}$ is equal to

\frac{2 a_{2}}{a_{2} + a_{3}}

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\frac{- 2 a_{2}}{a_{2} + a_{3}}

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\frac{2 a_{2}}{a_{1} + a_{3}}

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\frac{- 2 a_{2}}{a_{1} + a_{3}}

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Solution

The correct option is A $\frac{2 a_{2}}{a_{2} + a_{3}}$
Let the coefficients of four consecutive terms in the expansion be
$^{n} C_{r},^{n} C_{r + 1},^{n} C_{r + 2},^{n} C_{r + 3}$
So, $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}} + \frac{^{n} C_{r + 2}}{^{n} C_{r + 2} +^{n} C_{r + 3}} = \frac{^{n} C_{r}}{^{n + 1} C_{r + 1}} + \frac{^{n} C_{r + 2}}{^{n + 1} C_{r + 3}} = \frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2 (r + 2)}{n + 1}$
Check for option A.
$\frac{2 a_{2}}{a_{2} + a_{3}} = \frac{^{n} C_{r + 1} \times 2}{^{n} C_{r + 1} +^{n} C_{r + 2}} = \frac{2 \times^{n} C_{r + 1}}{^{n + 1} C_{r + 2}} = \frac{2 (r + 2)}{n + 1}$
Hence, option A is true. It can be shown, similarly that none of the other option is true.

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