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

If the coefficient of second,third and fourth terms in the expansion of $(1+x)^{2n}$ are in A.P.Show that $2n^2-9n+7=0$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
By the given condition we have $2nC_1,2nC_2,2nC_3$ are in A.P
$\therefore 2\times 2nC_2=2nC_1+2nC_3$
$\Rightarrow 2.\large\frac{2n(2n-1)}{1.2}=\frac{2n}{1}+\frac{2n(2n-1)(2n-2)}{1.2.3}$
$\Rightarrow 2n-1=1+\large\frac{(2n-1)(n-1)}{3}$
$\Rightarrow 6n-3=3n+2n^2-3n+1$
$\Rightarrow 2n^2-9n+7=0$
Hence proved.
answered Jun 24, 2014 by sreemathi.v
 

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
...