First slide

Binomial theorem for positive integral Index

Question

if a₁,a₂,a₃a,₄ are the coefficients of any four consecutive terms in the expansion of $(1 + x)^{n}$ , then $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}}$ is equal to

Moderate

A

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

$\frac{1}{2} \frac{a_{2}}{(a_{2} + a_{3})}$
C

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

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

Solution

Let a_1,a_2,a_3,a₄ be respectively the coefficients of
(r + 1)th, (r + 21)th, (r + 3)th and (r + 4)th terms in the
expansion of (1 + ,x)ⁿ .Then,
$a_{1} =^{n} C_{r}, a_{2} =^{n} C_{r + 1}, a_{3} =^{n} C_{r + 2}, a_{4} =^{'} C_{r + 3}$
Now $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{^{n} C_{r}}{^{n} C_{r} +^{n} C_{r + 1}} + \frac{^{n} C_{r + 2}}{^{n} C_{r + 2} +^{n} C_{r + 3}}$
$= \frac{^{n} C_{r}}{^{n + 1} C_{r + 1}} + \frac{^{n} C_{r + 2}}{n + 1_{C_{r + 3}}}$
$[∵^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}]$
$= \frac{^{n} C_{r}}{{\frac{n + 1}{r + 1}}^{n} C_{r}} + \frac{^{n} C_{r + 2}}{{\frac{n + 1}{r + 3}}^{n} C_{r + 2}} (∵^{n} C_{r} = {\frac{n}{r}}^{n - 1} C_{r - 1})$
$= \frac{r + 1}{n + 1} + \frac{r + 3}{n + 1} = \frac{2 (r + 2)}{n + 1}$
$= 2 \frac{^{n} C_{r + 1}}{^{n + 1} C_{r + 2}} = 2 \frac{^{n} C_{r + 1}}{^{n} C_{r + 1} +^{n} C_{r + 2}} = \frac{2 a_{2}}{a_{2} + a_{3}}$

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