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

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

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

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

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

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Solution

The correct option is C
$\frac{2 a_{2}}{a_{2} + a_{3}}$

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,
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{{^{n} C}_{r}}{{^{n} C}_{r} + {^{n} C}_{r + 1}} + \frac{{^{n} C}_{r + 2}}{{^{n} C}_{r + 2} + {^{n} C}_{r + 3}} = \frac{{^{n} C}_{r}}{{^{n + 1} C}_{r + 1}} + \frac{{^{n} C}_{r + 2}}{{^{n + 1} C}_{r + 3}} = \frac{{^{n} C}_{r}}{\frac{n + 1}{r + 1} {^{n} C}_{r}} + \frac{{^{n} C}_{r}}{\frac{n + 1}{r + 3} {^{n} C}_{r + 2}} = \frac{r + 1}{n + 1} + \frac{r + 3}{n + 3} = \frac{2 (r + 2)}{n + 1}$
Also, solving the R.H.S, we get
$\frac{2 a_{2}}{a_{2} + a_{3}} = 2 \frac{{^{n} C}_{r + 1}}{{^{n} C}_{r + 1} + {^{n} C}_{r + 2}} = 2 \frac{{^{n} C}_{r + 1}}{{^{n + 1} C}_{r + 2}} = \frac{2 (r + 2)}{n + 1}$

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