lf $a_{1}, a_{2}, a_{3}, a_{4}$ are the coefficients of $2^{n d}, 3^{r d}, 4^{t h}$ and $5^{t h}$ terms in the expansion of $(1 + x)^{n}$ respectively, 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{2 a_{2}}{a_{2} + a_{3}}

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

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\frac{4 a_{3}}{a_{2} + a_{3}}

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Solution

The correct option is C $\frac{2 a_{2}}{a_{2} + a_{3}}$
To find value of $(\frac{a_{1}}{a_{1} + a_{2}}) + (\frac{a_{3}}{a_{3} + a_{4}})$
$= \frac{1}{1 + \frac{a_{2}}{a_{1}}} + \frac{1}{1 + \frac{a_{4}}{a_{3}}}$ ...(i)
Now $a_{1} = n$
$a_{2} = \frac{n . (n - 1)}{2}$
$a_{3} = \frac{n . (n - 1) (n - 2)}{6}$
$a_{4} = \frac{n . (n - 1) (n - 2) (n - 3)}{24}$
Therefore, Eq(i) reduces to
$\frac{1}{1 + \frac{n - 1}{2}} + \frac{1}{1 + \frac{n - 3}{4}}$
$= \frac{2}{n + 1} + \frac{4}{n + 1}$
$= \frac{6}{n + 1}$
$= \frac{2.3}{3 + n - 2}$
$= \frac{2}{1 + \frac{n - 2}{3}}$
$= \frac{2}{1 + \frac{a_{3}}{a_{2}}}$
$= \frac{2. a_{2}}{a_{3} + a_{4}}$

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Similar questions

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

$\frac{a_{1}}{a_{1} + a_{2}}$ + $\frac{a_{3}}{a_{3} + a_{4}}$ =

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

Q. 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}}

If $a_{1}, a_{2}, a_{3}, a_{4}$ be the coefficeints of four consecutive terms in the expansion of $(1 + x)^{n}$ then prove that
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{(a_{3} + a_{4})} = \frac{2 a_{2}}{(a_{2} + a_{3})} .$

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 $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} =$

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lf a1,a2,a3,a4 are the coefficients of 2nd,3rd,4th and 5th terms in the expansion of (1+x)n respectively, then a1a1+a2+a3a3+a4

The correct option is C 2a2a2+a3To find value of (a1a1+a2)+(a3a3+a4)=11+a2a1+11+a4a3 ...(i)Now a1=na2=n.(n−1)2a3=n.(n−1)(n−2)6a4=n.(n−1)(n−2)(n−3)24Therefore, Eq(i) reduces to11+n−12+11+n−34=2n+1+4n+1=6n+1=2.33+n−2=21+n−23=21+a3a2=2.a2a3+a4

lf $a_{1}, a_{2}, a_{3}, a_{4}$ are the coefficients of $2^{n d}, 3^{r d}, 4^{t h}$ and $5^{t h}$ terms in the expansion of $(1 + x)^{n}$ respectively, then $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}}$