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

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
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

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