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# Binomial theorem watch

1. 4 part question

1(a)Find first 3 terms in ascending powers of x of binomial expansion (2-9x)^4
1st term: 16
2nd term: -288x
3rd term:1944x^2

(b) f(x)=(1+kx)(2-9x)^4 where k is constant
the expansion in ascending powers of x of f(x) up to and including the term x^2 is
A-232x+Bx^2 where A and B are constants

(1+kx)(16k-288x+1944x^2)
16k^2x-288kx^2+1944x^3k

this is my working out but it does not seem to lead anywhere any help much appreciated for part b and c and d.

(c)Find value of K

(d)Hence value for B
2. (Original post by Anonymous1502)
(1+kx)(16k-288x+1944x^2)
16k^2x-288kx^2+1944x^3k
Are you sure that you should have a k in the second bracket before you expand?
3. (Original post by Pangol)
Are you sure that you should have a k in the second bracket before you expand?
Not sure.
4. (Original post by Anonymous1502)
Not sure.
In your (correct) answer to part (a), you say that (2 - 9x)^4 = 16 - 288x + 1944x^2 + ...

So, in part (b), you should have that f(x) = (1 + kx)(2 - 9x)^4 = (1 + kx)(16 - 288x + 1944x^2 + ...)
5. (Original post by Pangol)
In your (correct) answer to part (a), you say that (2 - 9x)^4 = 16 - 288x + 1944x^2 + ...

So, in part (b), you should have that f(x) = (1 + kx)(2 - 9x)^4 = (1 + kx)(16 - 288x + 1944x^2 + ...)
So should I find the 4th term?
6. (Original post by Anonymous1502)
So should I find the 4th term?
You only need to compare this with A - 232x + Bx^2, so you only need the first three terms.
7. (Original post by Pangol)
You only need to compare this with A - 232x + Bx^2, so you only need the first three terms.
How do I compare?
8. (Original post by Anonymous1502)
How do I compare?
If they are the same, then then have the same constant term, the same coefficient of x, the same coefficient of x^2, etc.

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