binomial theorem - coefficients (1 Viewer)

[black_man]

would someone please be able to help me with identifying coefficients in binomial expansions. i'm not sure if others have encountered this aswell or i myself am making mistakes, but when a question comes up like this:

find the term independent of x in the expansion (x- 1/x^2)15

it seems as though equating the coefficients to 0 to find the term number will only work when the products are arranged in a certain way. in this case - 15Cr (x)^15-r and (-1)^r(1/x^2)r

in a number of questions similar to this i've attempted to solve it the other way round (with (x)^r and (1/x^2)^15-r. would this way still prove correct? i ask because in an exam i might be confronted with a question where i would get a different answer with each method and might get confused.

 

[FinalFantasy]

this is my way

"find the term independent of x in the expansion (x- 1/x^2)15"

First find general term:

15Ck *(x)^(15-k)(-1\x²)^k
=15Ck (-1)^k x^(15-k-2k)=nCk (-1)^k x^(15-3k)

For the term independant of x...
set 15-3k=0
hence k=5
so the term independant of x is
-15C5x^0

 

[black_man]

this is my way

"find the term independent of x in the expansion (x- 1/x^2)15"

First find general term:

15Ck *(x)^(15-k)(1\x²)^k
=15Ck x^(15-k-2k)=nCk x^(15-3k)

For the term independant of x...
set 15-3k=0
hence k=5
so the term independant of x is
15C5x^0

thankyou for your help. i did get that far, i wanted to ask whether it's possible to answer the question if you switched products, i.e addressed the question as 15Ck * (x)^k(1/x^2)^(15-k)

 

[black_man]

thankyou very much for your time, i think i have solved my problem

 

[FinalFantasy]

^of course you can!

15Ck *(x)^(k)(-1\x²)^(15-k)
=15Ck (-1)^(15-k) x^k x^(-30+2k)
=15Ck (-1)^(15-k) x^(3k-30)

term independant of x:
set 3k-30=0
k=10

hence term independant of x is
-15C10=-15C5

note i missed out a negative sign wen i did it da first time up there..

 

[acmilan]

Switching them makes no difference, this is because (x + 1/x²)¹⁵ = (1/x² + x)¹⁵