Binomial theorem Q (1 Viewer)

[karnage]

First and foremost, id like to say that binomial theorem is the hardest stuff ive ever done, seriously struggling

The following is from Exercise 5E Q7c of Camrbidge maths.

(x - 2y)^12

i. Find the coefficient with the greatest absolute value
ii. Find the term with the greatest absolute value if x = 2/3 and y = 3

Well doing the first bit

I did the usual: Tk+1 / Tk > 1 blah blah all the crap working out.

Then i simplied it to

-2y (13 - k) / xk > 1

Here am i supposed to make x and y = 1?

If so it comes down to

k > 26 which i am totally confused about.

Note:
I think my class is doing a different way to the way the TB explains.
We do Tk+1 = nCk, Tk = nCk-1

If anyone has done this one before, please shed your wisdom
I bet ya im doing it totally wrong.

 

[3unitz]

The following is from Exercise 5E Q7c of Camrbidge maths.

(x - 2y)^12

i. Find the coefficient with the greatest absolute value
ii. Find the term with the greatest absolute value if x = 2/3 and y = 3

i)
Tk+1 = (-1)^k . a^(n-k) . b^k . nCk
Tk+1 = (-1)^k . x^(12 - k) . (2y)^k . 12Ck
Tk+1 = (-1)^k . x^(12 - k) . (2^k). y^k . 12Ck

general coefficient = (2^k).12Ck

(2^k).12Ck / [2^(k-1)].12Ck-1 > 1
(2^k).12Ck > [2^(k-1)].12Ck-1
2.(12Ck) > 12Ck-1
2. 12!/[k!. (12 - k)!] > 12!/[(k-1)!. (12 - k + 1)!]
2 / [k!. (12 - k)!] > 1 /[(k-1)!. (12 - k + 1)!] (devide by through by 12!)
2 / (12 - k)! > k / (12 - k + 1)!] (multiply both sides by k!)
2 (12 - k + 1) > k (multiply both sides by (12 - k + 1)!)
24 - 2k + 2 > k
26 > 3k
8.6 > k
k = 8

greatest coefficient = 2^8.12C8 =126720

ii) same process but your general coefficient will be:
(2/3)^(12 - k) . (2^k). 3^k . 12Ck