A 1- a 2- a 3- a 4 are coefficients of any 4 consecutive terms in the expansion of 1- x n- then a 1- a 1- a 2- a 3- a 3- a 4-A- a 2 a 2- a 3B- 12 a 2 a 2- a 3C- 2 a 2 a 2-a 3D- 2 a 3 a 2- a 3

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Binomial Coefficients

a 1, a 2, a 3...

Question

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

a2a2+a3

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12a2a2+a3

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2a2a2+a3

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2a3a2+a3

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Solution

The correct option is C 2a2a2+a3
Let $a_{1}, a_{2}, a_{3}, a_{4}$ be respectively the coefficients of $(r + 1)^{t h}, (r + 2)^{t h}, (r + 3)^{t h}$ and $(r + 4)^{t h}$ 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,
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}} + \frac{^{n} C_{r + 2}}{^{n} C_{r + 2} +^{n} C_{r + 3}}$
$= \frac{^{n} C_{r}}{^{n + 1} C_{r + 1}} + \frac{^{n} C_{r + 2}}{^{n + 1} C_{r + 3}}$
$= \frac{^{n} C_{r}}{{\frac{n + 1}{r + 1}}^{n} C_{r}} + \frac{^{n} C_{r + 2}}{{\frac{n + 1}{r + 3}}^{n} C_{r + 2}}$ $[∵^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1}]$
$= \frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2 (r + 2)}{n + 1}$
$= 2 \times \frac{^{n} C_{r + 1}}{^{n + 1} C_{r + 2}} = 2 \times \frac{^{n} C_{r + 1}}{^{n} C_{r + 1} +^{n} C_{r + 2}}$
$= \frac{2 a_{2}}{a_{2} + a_{3}}$

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a1,a2,a3,a4 are coefficients of any 4 consecutive terms in the expansion of (1+x)n, then a1a1+a2+a3a3+a4=

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