Binomial expansion- Constant term(Question 2

Original post by realed

For this question i'm very confused about how you find the answer because i have no single digit of x https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FAdmissions%2FTMUA%2FPapers%2FTMUA%25202018%2520Paper%25202.pdf

You could (roughly) write a few of the terms down and see if you can spot which term it corresponds to. You could pull the 1/x^2 outside which might simplify it slightly.

(edited 9 months ago)

Reply 2

tonyiptony

11

Original post by realed

For this question i'm very confused about how you find the answer because i have no single digit of x https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FAdmissions%2FTMUA%2FPapers%2FTMUA%25202018%2520Paper%25202.pdf

Go for the freebies first (despite being a multiple choice and no marks for workings).
In this case, the freebie is "can you write down the binomial theorem for this particular problem".
The next one, slightly less freebie but still crucial, is "using above, can you locate the constant term".

Also I don't know what you mean by "no single digit of x", but probably not a very useful line.

Reply 3

Muttley79

20

Original post by realed

For this question i'm very confused about how you find the answer because i have no single digit of x https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FAdmissions%2FTMUA%2FPapers%2FTMUA%25202018%2520Paper%25202.pdf

post what you have tried

Reply 4

Ignorabimus

(a-b)^12 = sum from n=0 to 12 (-1)^n(12Cn)a^(12-n)b^n
so we want the constant term, which is x^0
so what a^(12-n)b^n will give that?
a=x^6,b=x^-2

Spoiler

(edited 9 months ago)

Reply 5

realed

OP

Original post by mqb2766

You could (roughly) write a few of the terms down and see if you can spot which term it corresponds to. You could pull the 1/x^2 outside which might simplify it slightly.

when i tried it i said 1/x^2= x^-2 but then i was like how on earth can this help me if i don't have a constant because you can't do binomial expansion without iy

Reply 6

Original post by realed

when i tried it i said 1/x^2= x^-2 but then i was like how on earth can this help me if i don't have a constant because you can't do binomial expansion without iy

If you have
(a+b)^n
then the expansion is
a^n + nC1*a^(n-1)b + nC2*a^(n-2)b^2 + .... + b^n
So when you multipy by "b"s the power of x gets reduced. Iif you wrote out a few of the terms you should see this. The question is asking for the coeffiicent of the term x^0.

Or alternatively factor out 1/x^(2) so
1/x^(24) (x^8 - 1)^12
and for the constant term youre looking for the term in (x^8 - 1)^12 which corresponds to x^24.

(edited 8 months ago)

Reply 7

realed

OP

if i factor x^-2 out i get (x^-2(X^8-1))^12

but what on earth happens next

Reply 8

Original post by realed

if i factor x^-2 out i get (x^-2(X^8-1))^12

but what on earth happens next

You factor x^(-2) out of each of the 12 terms. So the overall multiplier is x^(-24).
x^-24 * (X^8-1)^12

Then you pick the term corresponding to x^(24) in (X^8-1)^12, as when you multiply by x^(-24) it will give the constant term or the x^0 term.

(edited 8 months ago)

Reply 9

realed

OP

Original post by mqb2766

You factor x^(-2) out of each of the 12 terms. So the overall multiplier is x^(-24).
x^-24 * (X^8-1)^12

Then you pick the term corresponding to x^(24) in (X^8-1)^12, as when you multiply by x^(-24) it will give the constant term or the x^0 term.

wait so would that be (x^8-1)^3=-220 which is the answer as x^24 and x^-24 cancel out

Reply 10