Q7b(ii) - Fudging it. (1 Viewer)

[~ ReNcH ~]

For this question, I didn't have time to actually work it our properly, but I guessed it would be an "equating co-efficients" kind of proof. Was it?
Coz on the actual paper, I just said:
Co-eff. of x^k on LHS = whatever was on the exam paper
Co-eff. of x^k on RHS = ⁿC_k
.'. Equating co-efficients, LHS=RHS.

Was this the right "method". If it was, would my fudging get 1 mark??

 

[Rorix]

it was the correct method, from memory, so you should get the mark

 

[Gesus]

hehe i fudged the very last bit, i was actually on track, but running out of time and i had skipped a question, so just wrote out the answer and wrote "As required"
hehe
fingers crossed

 

[~ ReNcH ~]

it was the correct method, from memory, so you should get the mark

Hehe...hopefully.
Coz I don't think there's any way to tell whether you actually observed the expression first, before writing down the co-efficient on either side unless there's some intermediate step. So since it was only worth 1 mark, I hope I get it - to me, 1 mark counts for a lot.

 

[mojako]

I used the coeff of x^k but did sumthing wrong in the process.. I said coeff of x^k on the RHS in part (i) was "nCk - 1" and in the LHS its (n-1)C(k-1) + ... + (k)C(k-1). Then said since 1 = (k-1)C(k-1) we can move it to the LHS so we get the required identity in (ii).
.... hopefully they allow this..............
*faint hope...*

 

[~ ReNcH ~]

I used the coeff of x^k but did sumthing wrong in the process.. I said coeff of x^k on the RHS in part (i) was "nCk - 1" and in the LHS its (n-1)C(k-1) + ... + (k)C(k-1). Then said since 1 = (k-1)C(k-1) we can move it to the LHS so we get the required identity in (ii).
.... hopefully they allow this..............
*faint hope...*

Is my proof correct?
Coz I didn't actually look at the expression. I basically just wrote out the expression, and said "equating co-efficients" to make it look like I actually did do something.

 

[mojako]

Is my proof correct?
Coz I didn't actually look at the expression. I basically just wrote out the expression, and said "equating co-efficients" to make it look like I actually did do something.

yes it is
u r a very very lucky person

coeff of x^k on the LHS in part (i) is automatically the LHS in part (ii). There's no messing around to get that. So you don't need working. And so you're lucky!!
and I'm not!!

(sounds like I'm emotional )