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 _______

Correct answer: 0 ^ 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=(10)/(3) ....(iii) aligned & By (i) ( ii) & ^n C_1 ^n C_2 (x)/(y)=(9)/(2)...(iv) & By ( ii) ( iii) & ^n C_2 ^n C_3 (x)/(y)=9....(v) aligned aligned & By ( iv ) (v) & ^n C_1 ^n C_3 ^n C_2 ^n C_2 =(1)/(2) & (2 n^2(n-1)(n-2))/(6)=(n(n-1))/(2) (n(n-1))/(2) & 4 n-8=3 n-3 & n=5 & put in (v) aligned aligned & (x)/(y)=9 & x=9 y & put in (i) & ^5 C_1 x^4((x)/(9))=135 & x^5=27 × 9 & x=3, y=(1)/(3) & 6(n^3+x^2+y) & =6(125+9+(1)/(3)) & =806 aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Binomial Theorem

If the second, third and fourth terms in the expansion of (x+y)^n...

If the second, third and fourth terms in the expansion of

(x+y)^n

are 135,30 and

\frac{10}{3}

, respectively, then $6\left(n^3+x^2+y\right) is equal to _______

Correct answer

Step-by-step solution

{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \mathrm{y}=135

....(i)

{ }^n C_2 x^{n-2} y^2=30

....(ii)

{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^{\mathrm{n}-3} \mathrm{y}^3=\frac{10}{3}

....(iii)

\begin{aligned} & \text{ By } \frac{(i)}{(\text{ ii) }} \\ & \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)\end{aligned}

\begin{aligned} & \text{ By } \frac{(\text{ iv })}{(v)} \\ & \frac{{ }^n C_1{ }^n 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 \\ & \text{ put\text{ in } (v) }\end{aligned}

\begin{aligned} & \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\end{aligned}

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If 26((2^3)/(3) 12 2 + (2^5)/(5) 12 4 + (2^7)/(7) 12 6 + … + 2^ 13 13 12 12 ) = 3^ 13 - α… JEE Main 2026 If (1 - x^3)^ 10 = ∑ _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r , then 9a_9 a_ 10 is equal to… JEE Main 2026 If the coefficients of the middle terms in the binomial expansions of (1 + α x)^ 26 and… JEE Main 2026 The coefficient of x^2 in the expansion of (2x^2 + (1)/(x))^ 10 , x ≠ 0 , is : JEE Main 2026 If the sum of the coefficients of x^7 and x^ 14 in the expansion of ((1)/(x^3) - x^4)^n… JEE Main 2026 In the expansion of (9x- 1 3 x )^ 18 , x>0 , if the term independent of x is (221)k… JEE Main 2026 Let the smallest value of k ∈ ℕ , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 +… JEE Main 2026 If for 3 ≤ r ≤ 30 , 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals: JEE Main 2026 The number of elements in the set S = \ (r, k) : k ∈ ℤ and ^ 36 C_ r+1 = 6( ^ 35 C_r)… JEE Main 2026 Given below are two statements : Statement I: 25^ 13 +20^ 13 +8^ 13 +3^ 13 is divisible… JEE Main 2026 The sum of the coefficients of x^ 499 and x^ 500 in (1+x)^ 1000 +x(1+x)^ 999 +x^ 2 (1+x)^… JEE Main 2026 Let S=(1)/(25!)+(1)/(3!23!)+(1)/(5!21!)+… up to 13 terms. If 13 ~S= 2^ k n! , k ∈ ~N… JEE Main 2026 The value of ^ 100 C_ 50 51 + ^ 100 C_ 51 52 +….+ ^ 100 C_ 100 101 is : JEE Main 2026 The sum of all possible values of n ∈ N , so that the coefficients of x, x^ 2 and x^ 3 in… JEE Main 2026 Let C_ r denote the coefficient of x^ r in the binomial expansion of (1+x)^ n , n ∈ N, 0… JEE Main 2026 The coefficient of x^ 48 in (1+x)+2(1+x)^ 2 +3(1+x)^ 3 +…+100(1+x)^ 100 is equal to JEE Main 2026 Full Binomial Theorem list Binomial Theorem JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs