if in the expansion of $(1+x)^m (1-x)^n$, the coefficients of $x$ and $x^2$ are $3$ and $-6$. Find value of $m$ and $n$.

Question

If in the expansion of $(1+x)^m (1-x)^n$, the coefficients of $x$ and $x^2$ are $3$ and $-6$. Find value of $m$ and $n$.

Please help me solve the above problem

Answer 1

$(1+x)^m (1-x)^n = (1+mx+\binom m2x^2...)(1-nx+\binom n2 x^2...)$

$=1+(m-n)x+(\binom m2+\binom n2-mn)x^2...$,

so for this problem $m-n=3$ and $\binom m2 +\binom n2 - mn=-6. $

Can you solve for $m$ and $n$ now?

Answer 2

Hint:

Using Binomial theorem, you can expand each term.

The product of the expansions is shown as in here:

Now you have 2 equations in two variables $m$ and $n$. Solve to get $m=12$ and $n=9$.

Answer 3

You can use binomial coefficients as an approach. In $(x+1)^m$, the coefficient of the $x$ term is always $m \choose 1$, which is effectively equal to $m$.

Similarly, the coefficient for the $x$ term of $(1-x)^n$ is $-n$. So one of the $x^2$ terms will be $-mnx^2$ (the $x$'s and coefficients multiply together).

The coefficient of the $x^2$ term of $(x+1)^m$ is $m \choose 2$, which is the same as $\frac{m(m-1)}{2}$. Because $x^2$ is even, the coefficient for $(1-x)^n$ is $\frac{n(n-1)}{2}$. Assuming each $x^2$ term in one binomial expansion is multiplied by the $1$ on the end of the other binomial, you get the $x^2$ coefficient to be $$-mn + \frac{m(m-1) + n(n-1)}{2}$$ The coefficient of the $x$ term for $(x+1)^m$ is always $m$, and for $(1-x)^n$ is always $-n$. Thus, the coefficient for the $x$ term of the product both binomials, knowing that each respective $x$ term is multiplied by the $1$ on the other binomial, is $m-n$. Now plug in the initial values. $$m-n = 3$$ $$-mn + \frac{m(m-1) + n(n-1)}{2} = -6$$ $m=3+n$, so $$-3n-n^2 + \frac{(3+n)(2+n) + n(n-1)}{2} = -6$$ $$-3n-n^2 + \frac{6+5n+n^2 + n^2-n}{2} = -6$$ $$-3n-n^2 + 3+2n+n^2 = -6$$ $$-n + 3 = -6$$ $$n=9$$ Additionally, $m = 3+9 = 12$, so $(m,n) = (12,9)$

Answer 4

Let $$f(x)=(1+x)^m (1-x)^n= a_0 + a_1x + a_2x^2+....a_{m+n}x^{m+n}$$

We know that $$a_1= f'(0)=3$$ and $$a_2=f"(0)/2=-6$$

$$ f'(x) = m(1+x)^{m-1}(1-x)^n - n(1+x)^m(1-x)^{n-1}$$

Thus $$f'(0)= m-n=3$$

Also $$f"(x)= m(m-1)(1+x)^{m-2}(1-x)^n-m(1+x)^{m-1}(n)(1-x)^{n-1}-mn(1+x)^{m-1}(1-x)^{n-1}+n(n-1)(1+x)^m(1-x)^{n-2}$$

Thus $$f"(0)=(m-n)^2-(m+n)=-12$$

Resulting in $$m=12, n=9$$