7.Binomial Theorem
hard

જો $(x+y)^n$ ના વિસ્તરણમાં બીજા, ત્રીજા અને ચોથા પદો અનુક્રમે $135,30$ અને $\frac{10}{3}$ હોય, તો $6\left(n^3+x^2+y\right)=$ ...............

A

$305$

B

$806$

C

$604$

D

$204$

(JEE MAIN-2024)

Solution

$ { }^n C_1 x^{n-1} y=135 $ ………..($i$)

$ { }^n C_2 x^{n-2} y^2=30 $ …………($ii$)

$ { }^n C_3 x^{n-3} y^3=\frac{10}{3} $ …………($iii$)

$ \text { By } \frac{(i)}{(i i)} $

$ \frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2} $ …………($iv$)

$ \text { By } \frac{(\text { ii) }}{(\text { iii) }} $

$ \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{y}=9 $ ……………($v$)

$ \text { By } \frac{(\text { iv) }}{(v)} $

$ \frac{{ }^n C_1 C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2}$

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

$ 4 n-8=3 n-3 $

$ \Rightarrow n=5$

put in ($v$)

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

$ x=9 y $

$ \text { put in (i) } $

$ { }^5 C_1 x^4\left(\frac{x}{9}\right)=135 $

$ x^5=27 \times 9 $

$ \Rightarrow x=3, \quad y=\frac{1}{3} $

$ 6\left(n^3+x^2+y\right) $

$ =6\left(125+9+\frac{1}{3}\right) $

$ =806$

Standard 11
Mathematics

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