Binomial expansion help urgent?

Question

http://www.free-picture-host.com/viewimage.php?file=/images/FVUWcj1198925619.jpg
(nid urgent help can't do it)

I NEED HELP for the answer
NOT the formula
(in conclusion, i need double-checking)

Answer 1

Hi,

4096x^24 - 12,288x^21 + 16,896x^18 - 14,080x^15 + 7,920x^12 - 3,168x^9 + 924x^6 - 198x^3 + 30.9375 - 3.4375x^-3 + .2578125x^-6 - .01171875x^-9 + .000244140625x^-12

You can get this from your graphing calculator if you enter your coefficient for Y1, your first variable's exponent equation as Y2 and your second variable's exponent as Y3. In this case since both variables are "x". the Y2 and Y3 can be combined as Y2.

Start by re-writing the problem as (2x² - .5x^(-1))^12.

The coefficient equation is made of the following parts:

(binomial's exponent) nCr X where X allows this to vary from 0 to 12 for the 13 terms. This becomes the first part of our equation which is:
12 nCr X
This is multiplied by the first term's coefficient raised to the (binomial's exponent - X) times the second term's coefficient, including the "-" sign, raised to the X power. So, the entire expression is:
Y1 = 12 nCr X*(2)^(12-X)*(-.5)^X

The first variable's exponent is found from the first term's exponent times (binomial's exponent - X). This means that Y2 alone is Y2 = 2(12 - X).

The second variable's exponent is found from the second term's exponent times (X). This means that Y3 alone is Y3 = -1(X).

Since both variables are "x", these last 2 equations can be combined for this problem as Y2 = 2(12 - X) - X.

So the coefficient and exponent equations are:
Y1 = 12 nCr X*(2)^(12-X)*(-.5)^X <== coefficient
Y2 = 2(12 - X) - X <== exponent

To see these numbers, look at your TABLE starting at x = 0. When x = 0, Y1 is the coefficient and Y2 is the exponent for the first term. When x = 1, Y1 is the coefficient and Y2 is the exponent for the second term. When x = 2, Y1 is the coefficient and Y2 is the exponent for the first term, etc.

I hope this helps!! :-)

Answer 2

The row in Pascal's triangle corresponding to the 12th power is:
1 12 66 220 495 792 924 792 495 220 66 12 1

(2x - 1/(2x))^12

= 4096x^24 - 12*1024 x^21 + 66*256 x^18 - 220*64 x^15 + 495*16 x^12 - 792*4 x^9 + 924*x^6 - 792/4 * x^3 + 495/(16) - 220/(64) * x^(-3) + 66/(256) * x^(-6) - 12/(1024) * x^(-9) + 1/(2048)*x^(-12)

= 4096x^24 - 12288x^21 + 16896x^18 - 14080x^15 + 7920x^12 - 3168x^9 + 924x^6 - 198x^3 + 495/16 - 55/16 * x^(-3) + 33/128 * x^(-6) - 3/256 * x^(-9) + 1/4096 * x^(-12)

The power of the first term will be 2*12=24
The power of each successive term will drop by 3 (the power of x² drops by 1 and the power of 1/x increases by 1)
The last term will have a power of -12

Answer 3

(a + b)^12

use pascal's triangle to find the coefficients of each term, then replace a and b with 2x^2 and -1/2x
good luck