First slide

Binomial theorem for positive integral Index

Question

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ the coefficient of x and x² are 3 and -6 respectively, then m is

Moderate

A

6
B

9
C

12
D

24

Solution

$\begin{array}{r} (1 + x)^{m} (1 - x)^{n} = \{1 + mx + \frac{m (m - 1) x^{2}}{2!} + \dots\} [1 - nx + \frac{n (n - 1)}{2!} x^{2} - \dots] \end{array} = 1 + (m - n) x + [\frac{n^{2} - n}{2} - mn + \frac{(m^{2} - m)}{2}] x^{2} + \dots$
$\begin{aligned} Given, m - n = 3 \Rightarrow n = m - 3 \\ and \frac{n^{2} - n}{2} - mn + \frac{m^{2} - m}{2} = - 6 \\ \Rightarrow \frac{(m - 3) (m - 4)}{2} - m (m - 3) + \frac{m^{2} - m}{2} = - 6 \\ \Rightarrow m^{2} - 7 m + 12 - 2 m^{2} + 6 m + m^{2} - m + 12 = 0 \\ \Rightarrow - 2 m + 24 = 0 \Rightarrow m = 12 \end{aligned}$

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