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Question: If in the expansion of \({{\left( 1+x \right)}^{m}}{{\left( 1-x \right)}^{n}}\), the coefficients of...

If in the expansion of (1+x)m(1−x)n{{\left( 1+x \right)}^{m}}{{\left( 1-x \right)}^{n}}, the coefficients of x and x2{{x}^{2}} are 3 and - 6 respectively, then m is
(a) 6
(b) 9
(c) 12
(d) 24

Explanation

Solution

Hint: For solving this problem, first we expand both the terms by using binomial expansion. Now, we make possible pairs of coefficients and equate them as given in the problem. Using this methodology, we can easily obtain the answer.

Complete step-by-step answer:
According to the problem, we are given (1+x)m(1−x)n{{\left( 1+x \right)}^{m}}{{\left( 1-x \right)}^{n}} in which coefficient of x and x2{{x}^{2}} are 3 and -6. Now, we try to form two equations by using the coefficients of x and x2{{x}^{2}}.
The two useful expansion in binomial could be stated as:
(1+x)n=nC0+nC1x+nC2x2+nC3x3...............nCnxn (1−x)n=nC0−nC1x+nC2x2−nC3x3...............nCnxn \begin{aligned} & {{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{3}}{{x}^{3}}...............{}^{n}{{C}_{n}}{{x}^{n}} \\\ & {{\left( 1-x \right)}^{n}}={}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}-{}^{n}{{C}_{3}}{{x}^{3}}...............{}^{n}{{C}_{n}}{{x}^{n}} \\\ \end{aligned}
By using the above expansion, we can expand (1+x)m(1−x)n{{\left( 1+x \right)}^{m}}{{\left( 1-x \right)}^{n}} in simplified forms as:
(1+x)m(1−x)n=(mC0+mC1x+mC2x2+........mCmxm)(nC0−nC1x+nC2x2........nCnxn){{\left( 1+x \right)}^{m}}{{\left( 1-x \right)}^{n}}=\left( {}^{m}{{C}_{0}}+{}^{m}{{C}_{1}}x+{}^{m}{{C}_{2}}{{x}^{2}}+........{}^{m}{{C}_{m}}{{x}^{m}} \right)\left( {}^{n}{{C}_{0}}-{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}........{}^{n}{{C}_{n}}{{x}^{n}} \right)
Now, by observation we form pairs of multiplication which will give the coefficient of x:
mC0⋅(−nC1x)+(mC1x)⋅nC0=3x −mC0⋅nC1+mC1⋅nC0=3…(1) \begin{aligned} & {}^{m}{{C}_{0}}\cdot \left( -{}^{n}{{C}_{1}}x \right)+\left( {}^{m}{{C}_{1}}x \right)\cdot {}^{n}{{C}_{0}}=3x \\\ & -{}^{m}{{C}_{0}}\cdot {}^{n}{{C}_{1}}+{}^{m}{{C}_{1}}\cdot {}^{n}{{C}_{0}}=3\ldots (1) \\\ \end{aligned}
Similarly, pairs of multiplication for coefficient of x2{{x}^{2}} :
mC0⋅nC2x2+mC1x⋅(−nC1x)+mC2x2⋅nC0=−6x2 mC0⋅nC2−mC1⋅nC1+mC2⋅nC0=−6…(2) \begin{aligned} & {}^{m}{{C}_{0}}\cdot {}^{n}{{C}_{2}}{{x}^{2}}+{}^{m}{{C}_{1}}x\cdot \left( -{}^{n}{{C}_{1}}x \right)+{}^{m}{{C}_{2}}{{x}^{2}}\cdot {}^{n}{{C}_{0}}=-6{{x}^{2}} \\\ & {}^{m}{{C}_{0}}\cdot {}^{n}{{C}_{2}}-{}^{m}{{C}_{1}}\cdot {}^{n}{{C}_{1}}+{}^{m}{{C}_{2}}\cdot {}^{n}{{C}_{0}}=-6\ldots (2) \\\ \end{aligned}
Now we have two variables and two equations, so we can easily obtain our solution. Simplifying equation (1) and (2) and replacing n from equation (2) by using equation (1), we get

m−n=3 n=m−3 n(n−1)2−mn+m(m−1)2=−6 (m−3)(m−4)2−m(m−3)+m(m−1)2=−6 m2−7m+12−2m2+6m+m2−m2=−6 12−2m=−12 2m=24 m=242 m=12 \begin{aligned} & m-n=3 \\\ & n=m-3 \\\ & \dfrac{n\left( n-1 \right)}{2}-mn+\dfrac{m\left( m-1 \right)}{2}=-6 \\\ & \dfrac{\left( m-3 \right)\left( m-4 \right)}{2}-m\left( m-3 \right)+\dfrac{m\left( m-1 \right)}{2}=-6 \\\ & \dfrac{{{m}^{2}}-7m+12-2{{m}^{2}}+6m+{{m}^{2}}-m}{2}=-6 \\\ & 12-2m=-12 \\\ & 2m=24 \\\ & m=\dfrac{24}{2} \\\ & m=12 \\\ \end{aligned}
Therefore, the value of m is 12
Hence, option(c) is correct.

Note: The key concept involved in solving this problem is the knowledge of binomial expansion. Students must be careful while solving the equation because some silly mistakes might occur in calculation. Also, pairs of multiplication should be formed carefully by observation.