Let a=1+ ^2 C_2 3 ! + ^3 C_2 4 ! + ^4 C_2 5 ! + , b=1+ ^1 C_0+ ^1 C_1 1 ! + ^2 C_0+ ^2 C_1+ ^2 C_2 2 ! + ^3 C_0+ ^3 C_1+ ^3 C_2+ ^3 C_3 3 ! + Then (2 b)/(a^2) is equal to

Correct answer: 0 aligned & f(x)=1+ (1+x) 1 ! + (1+x)^2 2 ! + (1+x)^3 3 ! + . . & e^ (1+x) 1+x = 1 1+x +1+ (1+x) 2 ! + (1+x)^2 3 ! + (1+x)^2 4 ! & coef x^2 in RHS : 1+ ^2 C_2 3 + ^3 C_2 4 + =a aligned coeff. x^2 in L.H.S. e (1+x+(x^2)/(2 !)) (1-x+(x^2)/(2 !) ) is e-e+ e 2 ! =a aligned & b=1+(2)/(1 !)+(2^2)/(2 !)+(2^3)/(3 !)+ =e^2 & 2 ~b a^2 =8 aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2024 · 4 April · Morning Shift · Mathematics · Binomial Theorem

Let a=1+ ^2 C_2 3 ! + ^3 C_2 4 ! + ^4 C_2 5 ! + , b=1+ ^1 C_0+ ^1...

Let

a=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+\ldots

\mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+\ldots

Then

\frac{2 b}{a^2}

is equal to

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{f}(\mathrm{x})=1+\frac{(1+\mathrm{x})}{1 !}+\frac{(1+\mathrm{x})^2}{2 !}+\frac{(1+\mathrm{x})^3}{3 !}+\ldots . . \\ & \frac{\mathrm{e}^{(1+\mathrm{x})}}{1+\mathrm{x}}=\frac{1}{1+\mathrm{x}}+1+\frac{(1+\mathrm{x})}{2 !}+\frac{(1+\mathrm{x})^2}{3 !}+\frac{(1+\mathrm{x})^2}{4 !} \\ & \text{ coef } \mathrm{x}^2 \text{ in RHS : } 1+\frac{{ }^2 \mathrm{C}_2}{3}+\frac{{ }^3 \mathrm{C}_2}{4}+\ldots=\mathrm{a} \end{aligned}

coeff.

x^2

in L.H.S.

e

\left(1+x+\frac{x^2}{2 !}\right) \ldots

\left(1-x+\frac{x^2}{2 !} \ldots \ldots\right)

\mathrm{e}-\mathrm{e}+\frac{\mathrm{e}}{2 !}=\mathrm{a}

\begin{aligned} & \mathrm{b}=1+\frac{2}{1 !}+\frac{2^2}{2 !}+\frac{2^3}{3 !}+\ldots \ldots=\mathrm{e}^2 \\ & \frac{2 \mathrm{~b}}{\mathrm{a}^2}=8 \end{aligned}

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