JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Binomial Theorem · General Term and Middle Term
Answer$806$
Solution
Concept. Ratios of consecutive terms remove the powers and give $n$ and $\frac yx$. Formulas used. (1) $\dfrac{T_{r+1}}{T_r}=\dfrac{n-r+1}{r}\cdot\dfrac yx$ for $(x+y)^n$Step 1 - Two ratios. By (1), $$\frac{T_3}{T_2}=\frac{n-1}{2}\cdot\frac yx=\frac{30}{135}=\frac29,\qquad\frac{T_4}{T_3}=\frac{n-2}{3}\cdot\frac yx=\frac{10/3}{30}=\frac19.$$ Step 2 - Divide them. $$\frac{3(n-1)}{2(n-2)}=2\ \Longrightarrow\ 3n-3=4n-8\ \Longrightarrow\ n=5.$$ Then $2\cdot\frac yx=\frac29$, so $y=\frac x9$. Step 3 - Find $x,y$. $T_2=5x^4y=135$ gives $5x^4\cdot\frac x9=135$, so $x^5=243$, $x=3$, $y=\frac13$. Step 4 - Value. $$6\left(n^3+x^2+y\right)=6\left(125+9+\tfrac13\right)=750+54+2=806.$$ Answer: 806