if in the expansion of ( 1+x)^ m (1 -x) ^n, the coefficients of x and x^2 are 3 and -6 then n=?

Question

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

 

Answer 1

First off, note that (1+x)(1-x) = 1-x² . So any expansion of this would not contain an x term.

Secondly, since the coefficient of x is a positive number, we know that we have more (1+x) terms, or m>n.

 

So I can split up (1+x)^m into (1+x)^m-n (1+x)ⁿ I did this so that I can now multiply by the remaining factor to get

 

(1+x)^m-n (1-x²)ⁿ . The only way to get an x tem now is from the first factor. The only way to get a 3 as the coefficient of x is if m-n=3.

 

(1+x)³ = 1+3x+3x² +x³ (pascals triangle for the coefficients)

 

So your expression is equivalent to (1+3x+3x²+x³)(1-x²)ⁿ .

 

So now the coefficient of the x² term is -6. Thew only way to get an x² is by multiplying 1*the x² term in the expansion of the second factor added to 3x²*1. So the second factor must have a coefficient of -9 since -9+3=-6 to result in -6x².

 

So n must equal 9 to get the coefficient of -9x²  in the second expansion which means m=12.

 

So your expression must be (1+x)¹² (1-x)⁹.

 

Check

break into (1+x)³(1+x)⁹(1-x)⁹.........(1+3x+3x²+x³)(1-x²)⁹.....(1+3x+3x²+x³)(1-9x²+36x⁴.....) = 1+3x-6x²-26x³.....

Which contains the coefficients that you wanted.

 

Hope this somewhat helped.