If $x=3^{100},$ then $x$ has

Important Questions on Binomial Theorem

HARD

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2018>Q 7

${\left(\sqrt{3}+\sqrt{2}\right)}^4-{\left(\sqrt{3}-\sqrt{2}\right)}^4=$

MEDIUM

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2018>Q 8

If the sum of the coefficients in the expansions of ${\left(a^2{x}^2-2ax+1\right)}^{51}$ is zero, then $a$ is equal to

HARD

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2018>Q 9

If $5^{97}$ is divided by $52$, then the remainder obtained is

EASY

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2018>Q 10

$\frac{1}{9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!}$ Is equal to

EASY

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2018>Q 11

The remainder when $2^{2000}$ is divided by $17$ is

EASY

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2018>Q 12

$\left({}_{ }{}^7{C}_0+{}^7{C}_1\right)+\left({}^7{C}_2+{}^7{C}_3\right)+\dots +\left({}^7{C}_6+{}^7{C}_7\right)=$

EASY

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2020>Q 13

The middle term in the expansion of ${\left(1+\frac{1}{5}\right)}^{20}$ is

HARD

KEAM

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 15 - Binomial Theorem>KEAM 2019>Q 14

The third term in the expansion of ${\left(\frac{1}{x}+x^{{\log }_{10}x}\right)}^5,x>1$ is $1000,$ then $x$ equals.