If the coefficient of $4$ consecutive terms in the expansion of $(1 + x)^{n}$ are $a_{1}, a_{2}, a_{3}, a_{4}$ respectively , then show that:
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{2 a_{3}}{a_{2} + a_{3}}$ .

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Solution

To prove:- $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{2 a_{3}}{a_{2} + a_{3}}$
General term of the expansion ${(1 + x)}^{n}$ -
$T_{r + 1} =^{n} C_{r} {(1)}^{n - r} x^{r} =^{n} C_{r} x^{r}$
Let the $4$ consecutive terms be $T_{r + 1}, T_{r + 2}, T_{r + 3}$ and $T_{r + 4}$ respectively.
Therefore,
$a_{1} =^{n} C_{r}$
$a_{2} =^{n} C_{r + 1}$
$a_{3} =^{n} C_{r + 2}$
$a_{4} =^{n} C_{r + 3}$
Therefore,
$\frac{a_{1}}{a_{1} + a_{2}} = \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}} = \frac{(\frac{n!}{r! (n - r)!})}{(\frac{n!}{r! (n - r)!}) + (\frac{n!}{(r + 1)! (n - r - 1)!})} = \frac{(r + 1)}{n + 1}$
$\frac{a_{3}}{a_{3} + a_{4}} = \frac{^{n} C_{r + 2}}{^{n} C_{r + 2} +^{n} C_{r + 3}} = \frac{(\frac{n!}{(r + 2)! (n - r - 2)!})}{(\frac{n!}{(r + 2)! (n - r - 2)!}) + (\frac{n!}{(r + 3)! (n - r - 3)!})} = \frac{(r + 3)}{n + 1}$
$\frac{a_{3}}{a_{2} + a_{3}} = \frac{^{n} C_{r + 2}}{^{n} C_{r + 2} +^{n} C_{r + 1}} = \frac{(\frac{n!}{(r + 2)! (n - r - 2)!})}{(\frac{n!}{(r + 2)! (n - r - 2)!}) + (\frac{n!}{(r + 1)! (n - r - 1)!})} = \frac{(r + 2)}{n + 1}$
Now,
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{2 a_{3}}{a_{2} + a_{3}}$
$\frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2 (r + 2)}{n + 1}$
$2 r + 4 = 2 r + 4$
L.H.S. $=$ R.H.S.
Hence proved.

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