logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
0 votes

Expand the expression: $(x + \large\frac{1}{x}$$)^6$

$\begin{array}{1 1}x^6 + 6x^4 + 15x^2 + 20 + 15\large\frac{1}{x^2} + 6\large\frac{1}{x^4} + \large\frac{1}{x^6} \\ x^6 + 6x^4 + 15x^2 + 15\large\frac{1}{x^2} + 6\large\frac{1}{x^4} + \large\frac{1}{x^6} \\ x^6 + 5x^4 + 20x^2 + 25 + 5\large\frac{1}{x^2} + 20\large\frac{1}{x^4} + \large\frac{1}{x^6} \\ x^6 + 15x^4 + 6x^2 + 20 + 6\large\frac{1}{x^2} + 15\large\frac{1}{x^4} + \large\frac{1}{x^6}\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
Given $(x + \large\frac{1}{x}$$)^6 \rightarrow a = x$ and $b = \large\frac{1}{x}$
$\Rightarrow (x + \large\frac{1}{x}$$)^6 = \large \sum \limits_{k=0}^{6}\; $$^6 \large C$$_k \; x^{6-k}(1/x)^k $
$\qquad = \; ^6 \large C$$_0 \; x^{6-0}(1/x)^0 + \; ^6 \large C$$_1 \; x^{6-1}(1/x)^1 + \; ^6 \large C$$_2 \; x^{6-2}(1/x)^2+ \; ^6 \large C$$_3 \; x^{6-3}(1/x)^3 + \; ^6 \large C$$_4 \; x^{6-4}(1/x)^4 + \; ^6 \large C$$_5 \; x^{6-5}(1/x)^5 + \; ^6 \large C$$_6 \; x^{6-6}(1/x)^6$
$\qquad = x^6 + 6x^5\large\frac{1}{x}$$ + 15x^4\large\frac{1}{x^2}$$+20x^3\large\frac{1}{x^4}$$+6x\large\frac{1}{x^4}$$+\large\frac{1}{x^6}$
$\qquad = x^6 + 6x^4 + 15x^2 + 20 + 15\large\frac{1}{x^2}$$ + 6\large\frac{1}{x^4} $$ + \large\frac{1}{x^6}$
answered Apr 9, 2014 by balaji.thirumalai
edited Apr 9, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...