Binomial theorem (1 Viewer)

[denoz]

Find the coefficient of y^-3 in:

(y+1/y)^10 (y-1/y)^7

i could do this if i expanded it out but it would take to long. So is there any alternate method i could use to make it quicker.


And also in questions like (2x-3y)^6 where you have to find the greatest coefficient. when you sub it into :

T(k+1)/T(k) = (n-k+1)/k . b/a

When i put in b=-3y i always seem to get an invalid answer but when i sub in b=3y it works out alrite. Is this what i am suppose to do?

 

[Affinity]

Quicker erh... sort of.. but it's a bit ad hoc:
(y+1/y)^10 (y-1/y)^7 = (y^2 - 1/y^2)^7 (y+1/y)^3

the first bracket has powers (14,10,6,2,-2,-6,-10,-14) and the 2nd one has
(3,1,-1,-3) so the only ways to get a power of -3 is to match -2 with -1 and 3 with -6

so it becomes (3 C 2) * (7 C 4) + (3 C 0) * (7 C 5)


2nd Question: Hmm not quite.. you do substitute 3y instead of -3y BUT, you will need check if that term is POSITIVE. If it is, then fine. else that's not the greatest term and you will need to look at the previous or the next term.

The problem here is that with the negative sign in (2x-3y), the terms alternate from positive to negative.

 

[denoz]

(y+1/y)^10 (y-1/y)^7 = (y^2 - 1/y^2)^7 (y+1/y)^3

how did you do this step?

 

[pLuvia]

The difference between two squares.

Since (y+1/y)*(y-1/y)=(y²+1/y²). And so he used,
(y-1/y)⁷(y+1/y)⁷(y+1/y)³

Splitting the (y+1/y)¹⁰ into (y+1/y)⁷(y+1/y)³

=(y² - 1/y²)⁷(y+1/y)³

 

[denoz]

I see ... thanks