- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Evaluate: $\frac{a^{2 n+1} \times a^{(2 n+1)(2 n-1)}}{a^{n(4 n-1)}\times(a^{2})^{2 n+3}}$.
Given: $\frac{a^{2n+1} \times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times(a^{2})^{2n+3}}$.
To do: To solve: $\frac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times(a^{2})^{2n+3}}$.
Solution:
$\frac{a^{2n+1}\times a^{(2n+1)(2n-1)}}{a^{n(4n-1)}\times(a^{2})^{2n+3}}$
$=\frac{a^{( 2n+1)} \times a^{( 4n^2-1)}}{a^{(4n^2-n)}\times a^{( 4n+6)}}$
$=\frac{a^{( 2n+1+4n^2-1)}}{a^{( 4n^2-n+4n+6)}}$
$=a^{( 2n+1+4n^2-1-4n^2+n-4n-6)}$
$=a^{( -n-6)}$
$=a^{-( n+6)}$
$=\frac{1}{a^{( n+6)}}$
$=\frac{1}{a^n\times a^{6}}$
- Related Articles
- Prove that:\( \frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}=\frac{3}{2} \)
- Simplify the following:\( \frac{5 \times 25^{n+1}-25 \times 5^{2 n}}{5 \times 5^{2 n+3}-(25)^{n+1}} \)
- Solve: $\frac{n-1}{2}-\frac{n-2}{3}=\frac{n-4}{7}$.
- Simplify the following:\( \frac{5^{n+3}-6 \times 5^{n+1}}{9 \times 5^{n}-2^{2} \times 5^{n}} \)
- Verify: \( 1+2+3+\ldots+n=\frac{n(n+1)}{2} \), taking \( n=6 \) and \( 15 . \)
- Evaluate: \( 27^{-\frac{2}{3}} \times 9^{-\frac{1}{2}} \p 81^{-\frac{5}{4}} \)"\n
- Simplify the following:\( \frac{6(8)^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(8)^{n}} \)
- Find the value of $4m^2 n^2 \times 7m^3$ when $m= \frac{-1}{2}$ and $n= 4$.
- Simplify the following:\( \frac{3^n \times 9^{n+1}}{3^{n-1} \times 9^{n-1}} \)
- Python Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- C++ Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! + …… n/n!
- Java Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- Find (1^n + 2^n + 3^n + 4^n) mod 5 in C++
- Prove the following by using the principle a mathematical induction for all $n\epsilon N$.$1+3+3^{2}+\ldots+3^{n-1}=\frac{\left(3^{n}-1\right)}{2}$
- Find the sum:\( 4-\frac{1}{n}+4-\frac{2}{n}+4-\frac{3}{n}+\ldots \) upto \( n \) terms

Advertisements