Binomial theorem help needed thks
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Binomial theorem help needed thks

[From: ] [author: ] [Date: 11-09-23] [Hit: ]
.(1+x)^m = 1 + m*x + (m*(m-1))/2 * (x^2) + ...So when you multiply two such terms (1+x)^m and (1-x)^n, then it gets complicated.......
if in the expansion of (1+x)^m * (1-x)^n, the coefficients of x and x^2 are 3 and -6 respectively, find the value of m.

Pls show steps
btw m=12

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In the binomial expansion, you have the following property:
(1+x)^m = 1 + (mC1)*x + (mC2)*(x^2) + ...
(1+x)^m = 1 + m*x + (m*(m-1))/2 * (x^2) + ...

So when you multiply two such terms (1+x)^m and (1-x)^n, then it gets complicated.

But here, you are going to worry only about the coefficients of x and x^2. Now if you think, you will see that you will get 'x' terms only when the constant on one side multiplies with the 'x' term of the other side. Hence it will happen with mx multiplies with the constant 1 (from the power n term) and when -nx multiplies with the constant 1 (from the power m term). So you have the coefficient of x as:
x: m - n = 3
This is one equation that links m and n.

Now we have the x^2 term. And it is slightly trickier. You get (x^2) when you multiply 1 with a (x^2) term or if you multiply an x term with another x term. For the direct (x^2) terms, you have the ones that were listed out previously: (m*(m-1))/2 and (n*(n-1))/2. For the product of the two x terms you have (mx) and (-nx). So in total you have:

x^2: (m*(m-1))/2 + (n*(n-1))/2 - mn = -6
Simplifying this and plugging in the previous relation yields:
n = 9
m = n + 3 = 9 + 3 = 12

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x is a real number
1
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