Querying Binomial Q from last assessmnet (1 Viewer)

[vds700]

The question is:

Find the term independent of x in the expansion of (x^2 - (1/x))^6.

What i did was

T_r+1 = 6Cr(x^2)^(6-r)(-1/x)^r
=-6Cr x^(12-3r)
for constant,

12 - 3r = 0
r = 4
therefore constant term = -6C4 = -15

worth 2 marks, i only got 1 though.

In the marking criteria, they do not have the minus in the T_r+1, so their answer is 15.

There is a very similar example in Fitzpatrick, find the constant term of (2x^3 -(1/x))^12. He put a minus in, so the genral term is 12Cr(2x^3)^(12-r) . (-1/x)^r.

Is what I have done correct?

 

[tacogym27101990]

what i've always done if theres a question like that,
i treat it as (x²+(-1/x))⁶

so then you have ⁶C_k. [x²]^6-k.(-1)^k/(x)^k

then once you find k if k is even, you know it will be a positive coefficient

so i would say what you've done is incorrect, but hey ure much smarter than me

 

[vds700]

damn i didn't see that. grrr 1 mark off 100%, stupid brain.

Thanks for your input

 

[lyounamu]

damn i didn't see that. grrr 1 mark off 100%, stupid brain.

Thanks for your input

Don't worry. A lot of people in my class missed that. Justin got one mark off 100% as well.