Expert: Paul Klarreich - 10/10/2009

Question

find the coefficient of the term independent of x in the expansion (1+x+2x^3)(3/2x^2-1/3x)^9 also
if the coefficient of x^7 in (ax^2+ 1/bx)^11 is equal to the coefficient of x^-7 in (ax^2 - 1/bx)^11, then show that ab=1


Answer
Questioner: Sneha
Country: India
Category: Calculus
Private: No
Subject: binomial theorem
Question:
find the coefficient of the term independent of x in the expansion (1+x+2x^3)(3/2x^2-1/3x)^9 also
if the coefficient of x^7 in (ax^2+ 1/bx)^11 is equal to the coefficient of x^-7 in (ax^2 - 1/bx)^11, then show that ab=1
.............................................

Hi, Sneha,

Since you didn't indicate what you tried to do, I assume you just need to get started, and you already know the basics of the Binomal Expansion.

(1+x+2x^3)(3/2x^2-1/3x)^9

You did not indicate what you meant, so I think you meant:

( (3/2)x^2 - 1/(3x) )^9

because otherwise the problem makes no sense

[next time, parenthesize more carefully -- it is not really my job to figure out your question, just to answer]

Expanding, the first term has  x^18, then they go down by 3.  So the 7th term has x^0, i.e. a constant.  When that is hit by '1' you still have a constant.

Also, the 8th term has x^-3.  When that is hit by 2x^3, you get another constant.

Figure those terms, add them and you have you answer.

...................................

(ax^2+ 1/bx)^11  means (ax^2+ 1/(bx))^11, I assume.

Now the terms start at x^22, go down by 3, and reach


if the coefficient of x^7 in (ax^2+ 1/bx)^11 is equal to the coefficient of
x^-7 in (ax^2 - 1/bx)^11, then show that ab=1


(ax^2+ 1/bx)^11  means (ax^2+ 1/(bx))^11, I assume.

Now the terms in (ax^2+ 1/(bx))^11 start at x^22, go down by 3, and reach x^7 at the 6th term.  Compute that.

(ax^2 - 1/bx)^11 has terms starting at  x^22.  Now the exponents will be:

22, 19, 16, 13, 10, 7, 4, 1, -2, -5, -8..  oops, no  x^-7.  Why???