If the second, third and fourth terms in the expansion of ( x + y ) n are 135, 30, and 10 3 , respectively, then 6 ( n 3 + x 2 + y ) is equal to ______. [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

Ans.
(806)

$T_{r + 1} = {}^{n}C_{r} x^{n - r} y^{r}$

$∴$    ${}^{n}{C_{1}} x^{n - 1} y = 135,$ ...(i)

   ${}^{n}{C_{2}} x^{n - 2} y^{2} = 30$ ...(ii)

and ${}^{n}{C_{3}} x^{n - 3} y^{3} = \frac{10}{3}$ ...(iii)

By (i) and (ii), we have

   $\frac{{}^{n}{C_{1}} x}{{}^{n}{C_{2}} y} = \frac{135}{30} = \frac{9}{2}$ ...(iv)

By (ii) and (iii), we have

$\frac{{}^{n}{C_{2}} x}{{}^{n}{C_{3}} y} = \frac{30}{\frac{10}{3}} = 9$ ...(v)

By (iv) and (v), we have

$\frac{{}^{n}{C_{1}} x}{{}^{n}{C_{2}} y} \times \frac{{}^{n}{C_{3}} y}{{}^{n}{C_{2}} x} = \frac{\frac{9}{2}}{9}$

$\Rightarrow \frac{{}^{n}{C_{1} {}^{n}{C_{3}}}}{{({}^{n}{C_{2}})}^{2}} = \frac{1}{2} \Rightarrow 2 {}^{n}{C_{1}} {}^{n}{C_{3}} = {({}^{n}{C_{2}})}^{2}$

$\Rightarrow \frac{2 n n!}{3! (n - 3)!} = {(\frac{n!}{2! (n - 2)!})}^{2}$

$\Rightarrow \frac{2 n^{2} (n - 1) (n - 2)}{6} = {(\frac{n (n - 1)}{2})}^{2}$

$\Rightarrow \frac{n^{2} (n - 1) (n - 2)}{3} = \frac{n^{2} {(n - 1)}^{2}}{4} \Rightarrow \frac{n - 2}{3} = \frac{n - 1}{4}$

$\Rightarrow 4 n - 8 = 3 n - 3 \Rightarrow n = 5$

$\Rightarrow \frac{{}^{5}{C_{2} x}}{{}^{5}{C_{3}} y} = 9$ [Using (v)]

$\Rightarrow \frac{x}{y} = 9$

$\Rightarrow x = 9 y$ ...(vi)

$∴$ From (i), ${}^{5}{C_{1}} x^{4} \frac{x}{9} = 135 \Rightarrow x^{5} = \frac{135 \times 9}{5}$

$\Rightarrow x^{5} = 3^{5}$

$\Rightarrow x = 3$ and $y = \frac{1}{3}$ [Using (vi)]

Hence, $6 (n^{3} + x^{2} + y) = 6 (125 + 9 + \frac{1}{3}) = 806$

Want Us to Email you About Special Offers & Updates?