[SOLVED] Binomial Theorem question...

• Feb 1st 2008, 06:21 AM
looi76
[SOLVED] Binomial Theorem question...
Question:
Expand (1-3x)^10 up to and including the term x². Deduce the coefficient of x² in the expansion of (1+3x)^2(1-3x)^10.

Attempt:
(1-3x)^10 = 10C0*1^10 + 10C1*1^9*(-3x) + 10C2*1^8*(-3x)^2 +.....
(1-3x)^10 = 1*1 + 10*1*(-3x) + 45*1*(-3x)^2 +.....
(1-3x)^10 = 1 - 30x + 405x² +.....

(1+3x)^2 = 2C0*1^2 + 2C1*1*(3x) + 2C2*(3x)²
(1+3x)^2 = 1*1 + 2*1*3x + 1*9x²
(1+3x)^2 = 1 + 6x + 9x²

= (1 + 6x + 9x²)(1 - 30x + 405x² +.....)
Coefficient of x² = 405 + 180 + 9
Coefficient of x² = 594

Answer in textbook is 234, what is wrong?
• Feb 1st 2008, 06:26 AM
colby2152
Quote:

Originally Posted by looi76
= (1 + 6x + 9x²)(1 - 30x + 405x² +.....)
Coefficient of x² = 405 + 180 + 9
Coefficient of x² = 594

Answer in textbook is 234, what is wrong?

I didn't check the beginning, but I know you went wrong here...

$(1 + 6x + 9x^2)(1 - 30x + 405x^2) =$
$= 1 - 30x + 405x^2 - 30x - 180x^2 + 6(405)x^3 + 9x^2 - 270x^3 + 9(405)x^4$

$(405-180+9)x^2 = 234x^2$