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In the expansion of $(1+x)^m.(1-x)^n, $ the coefficients of $x$ and $x^2$ are $3$ and $-6$ respectively, then $m=?$

$\begin{array}{1 1} 6 \\ 9 \\ 12 \\ 24 \end{array} $

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$(1+x)^m.(1-x)^n=$
$\bigg[1+mx+\large\frac{m(m-1)}{2!}$$x^2+....\bigg]\times\bigg[1-nx+\large\frac{n(n-1)}{2!}$$x^2+......\bigg]$
$\therefore\:$ coeff. of $x=m-n$ and
Coeff. of $x^2=\large\frac{m(m-1)}{2}$$+\large\frac{n(n-1)}{2}$$-mn$
Given: $ m-n=3$ and
$\large\frac{m(m-1)}{2}$$+\large\frac{n(n-1)}{2}$$-mn=-6$
Solving these two equations we get $m=12$
answered Oct 12, 2013 by rvidyagovindarajan_1
 

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