If a1,a2,a3,a4 are the coefficients of any four consecutive terms in the ex

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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}} =

Options:

A . a2a2+a3

B . 12 a2a2+a3

C . 2a2a2+a3

D . 2a3a2+a3

Answer: Option C
:
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}} = \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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