Let a =1+frac 2 C 2 3! +frac 3 C 2 4! +frac 4 C 2 5! +ldots b =1+frac 1 C 0 + 1 C 1 1! +frac 2 C 0 + 2 C 1 + 2 C 2 2! +frac 3 C 0 + 3 C 1 + 3 C 2 + 3 C 3 3! +ldots Then frac 2b a 2 is equal to \!\! \!\!

Mathematics · Binomial Theorem

JEE Main 2024 — 4 April, Shift 1 — Question 26

Let $\text{a}=1+\frac{{{~}^{2}}{{\text{C}}_{2}}}{3!}+\frac{{{~}^{3}}{{\text{C}}_{2}}}{4!}+\frac{{{~}^{4}}{{\text{C}}_{2}}}{5!}+\ldots$

$\text{b}=1+\frac{{{~}^{1}}{{\text{C}}_{0}}+{{~}^{1}}{{\text{C}}_{1}}}{1!}+\frac{{{~}^{2}}{{\text{C}}_{0}}+{{~}^{2}}{{\text{C}}_{1}}+{{~}^{2}}{{\text{C}}_{2}}}{2!}+\frac{{{~}^{3}}{{\text{C}}_{0}}+{{~}^{3}}{{\text{C}}_{1}}+{{~}^{3}}{{\text{C}}_{2}}+{{~}^{3}}{{\text{C}}_{3}}}{3!}+\ldots$

Then $\frac{2b}{{{a}^{2}}}$ is equal to $\text{ }\!\!~\!\!\text{ }$

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{f}(\mathrm{x})=1+\frac{(1+\mathrm{x})}{1!}+\frac{(1+\mathrm{x})^{2}}{2!}+\frac{(1+\mathrm{x})^{3}}{3!}+\ldots$ .

$\frac{e^{(1+x)}}{1+x}=\frac{1}{1+x}+1+\frac{(1+x)}{2!}+\frac{(1+x)^{2}}{3!}+\frac{(1+x)^{2}}{4!}$

$\operatorname{coeff} \mathrm{x}^{2}$ in RHS : $1+\frac{{ }^{2} \mathrm{C}_{2}}{3}+\frac{{ }^{3} \mathrm{C}_{2}}{4}+\ldots=\mathrm{a}$

coeff. $x^{2}$ in L.H.S. $e\left(1+x+\frac{x^{2}}{2!}\right) \ldots\left(1-x+\frac{x^{2}}{2!} \ldots \ldots\right)$ is $e-e+\frac{e}{2!}=a$

$\mathrm{b}=1+\frac{2}{1!}+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\ldots \ldots=\mathrm{e}^{2}$

$\frac{2 \mathrm{~b}}{\mathrm{a}^{2}}=8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Series involving sum & product of Binomial Coefficients

The sum of all rational terms in the expansion of (2^1/5+5^1/3 )^15\\ is equal to:

Answer: 8

Step-by-step solution

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