First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

6
B

9
C

12
D

24

Solution

We have,
$(1 + x)^{m} (1 - x)^{n} = (^{m} C_{0} +^{m} C_{1} x +^{m} C_{2} x^{2} + \dots +^{m} C_{m} x^{m}) \times (^{n} C_{0} -^{n} C_{1} x +^{n} C_{2} x^{2} \dots . . + {(- 1)^{n}}^{n} C_{n} x^{n})$
$=^{m} {C_{0}}^{n} C_{0} - (^{m} {C_{0}}^{n} C_{1} -^{n} {C_{0}}^{m} C_{1}) x + (^{m} {C_{0}}^{n} C_{2} +^{n} {C_{0}}^{m} C_{2} -^{m} C_{1} \cdot^{n} C_{1}) x^{2} + \dots . .$
It is given that the coefficients of $x$ and $x^{2}$ in the expansion
$(1 + x)^{m} (1 - x)^{m}$ are 3 and -6 respectively
$∴ - (^{m} {C_{0}}^{n} C_{1} -^{n} {C_{0}}^{m} C_{1}) = 3$
and , $^{m} {C_{0}}^{n} C_{2} +^{n} {C_{0}}^{m} C_{2} -^{m} {C_{1}}^{n} C_{1} = - 6$
$\Rightarrow m - n = 3$ and $n (n - 1) + m (m - 1) - 2 m n = - 12$
$\Rightarrow m - n = 3$ and $(m - n)^{2} - (m + n) = - 12$
$\Rightarrow m - n = 3$ and $m + n = 21$
$\Rightarrow m = 12, n = 9$

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