# Binomial theorem question

1. Jan 14, 2006

### preet

I'm having problems figuring out how to do part (b) of this question.

a) expand $$(1-2x)^3$$ and $$(1+1/x)^5$$
b) Find, in the expansion of $$(1-2x)^3$$$$(1+1/x)^5$$
i) the constant term
ii) the coeffecient of x

I've done part a, and I know the formula for a general term for an expansion of a binomial. I don't get what I do in part b though..

TiA
Preet

2. Jan 14, 2006

### EnumaElish

When you expand the two power expressions in (a) then multiply them, what will you get as the constant term? (The answer isn't 1 because the x and the 1/x terms of the same power will cancel each other.) Then, what is the coefficient of x? Again you have to be careful because of cancellations.

Example: If both expressions were of power 1, i.e. (1-2x)(1+1/x), then:
(1-2x)(1+1/x) = 1 + 1/x - 2x - 2x/x = -1 + 1/x - 2x. So (i) -1 and (ii) -2.

3. Jan 14, 2006

### preet

I was thinking there would be a more sensible way to do it since Id have to expand (4 terms)(6 terms).

4. Jan 15, 2006

### HallsofIvy

Staff Emeritus
You don't have to multiply out the whole thing.
(1- 2x)3 involves terms with x to the 0, 1, 2, 3 powers and you have already found the coefficients.
(1+ 1/x)5 involves terms with x to the 0, -1, -2, -3, -4, -5 powers and you have already found the coefficients.

I will use (n, m) to mean "the terms in (1-2x)3 with exponent n and the term in (1+1/x)5 with exponent m".

The "constant", with exponent 0, requires multiplying the coefficients of the (0, 0), (1, -1), (2, -2), and (3, -3) powers together then adding them.

The "x" term, with coefficient 1, requires multiplying the coefficients of the (1, 0), (2, -1), (3, -2) powers together and then adding them.

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