If in the expansion of {(1 + x)^m}{(1 - x)^n} | vedclass

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Question

(c) ${(1 + x)^m}{(1 - x)^n}$ $ = \left( {1 + mx + \frac{{m(m - 1){x^2}}}{{2!}} + ....} \right)\,\left( {1 - nx + \frac{{n(n - 1)}}{{2!}}{x^2} - ....}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(c) ${(1 + x)^m}{(1 - x)^n}$ $ = \left( {1 + mx + \frac{{m(m - 1){x^2}}}{{2!}} + ....} \right)\,\left( {1 - nx + \frac{{n(n - 1)}}{{2!}}{x^2} - ....} \right)$

$ = 1 + (m - n)x + \left[ {\frac{{{n^2} - n}}{2} - mn + \frac{{({m^2} - m)}}{2}} \right]{x^2}$+........

Given, $m -n = 3$ or $n = m -3$ 

Hence $\frac{{{n^2} - n}}{2} - mn + \frac{{{m^2} - m}}{2} = - 6$

==> $\frac{{(m - 3)(m - 4)}}{2} - m(m - 3) + \frac{{{m^2} - m}}{2} = - 6$

==> ${m^2} - 7m + 12 - 2{m^2} + 6m + {m^2} - m + 12 = 0$

==> $ - 2m + 24 = 0\,\,\, \Rightarrow m = 12$

If in the expansion of ${(1 + x)^m}{(1 - x)^n}$, the coefficient of $x$ and ${x^2}$ are $3$ and $-6$ respectively, then m is

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