logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

If the sum of odd numbered terms and even numbered terms in the expansion of $(x+a)^n$ are $A,B$, then $(x^2-a^2)^n=?$

$\begin{array}{1 1} A^2 - B^2 \\ A^2+B^2 \\ 4AB \\ (A+B)^2 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
$(x+a)^n=^nC_0x^n+^nC_1x^{n-1}a+..............$
$=(^nC_0x^n+^nC_2x^{n-2}a^2+..........)+(^nC_1x^{n-1}a+^nC_3x^{n-3}a^3+.....)$
Given : $A=^nC_0x^n+^nC_2x^{n-2}a^2+...............$
$B=^nC_1x^{n-1}.a+^nC_3x^{n-3}a^3+............$
$\Rightarrow\:(x+a)^n=A+B$
$(x-a)^n=^nC_0x^n-^nC_1x^{n-1}a+^nC_2x^{n-2}a^2-.............$
$=(^nC_0x^n+^nC_2x^{n-2}a^2+.........)-(^nC_1x^{n-1}a+^nC_3x^{n-3}a^3+....)$
$\Rightarrow\:(x-a)^n=A-B$
$\therefore\:(x^2-a^2)^n=(x+a)^n.(x-a)^n$
$=(A+B)(A-B)=A^2-B^2$
answered Sep 24, 2013 by rvidyagovindarajan_1
 

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
...