Is My Attempt at Solving the Binomial Expansion Homework Correct?

[bllnsr]

Homework Statement

http://i47.tinypic.com/29o3ehc.png

Homework Equations

(1+x)^n=[1+nx/1!+(n)(n-1)x^2/2!+...+b^n]

The Attempt at a Solution

here's my attempt to part (i)
http://i47.tinypic.com/2yv6y3s.png
is it correct?

 

[Dick]

Homework Statement

http://i47.tinypic.com/29o3ehc.png

Homework Equations

(1+x)^n=[1+nx/1!+(n)(n-1)x^2/2!+...+b^n]

The Attempt at a Solution

here's my attempt to part (i)
http://i47.tinypic.com/2yv6y3s.png
is it correct?

Looks ok so far.

 

[skiller]

here's my attempt to part (i)
http://i47.tinypic.com/2yv6y3s.png
is it correct?

You got the correct answer somehow, but your working is flawed. Can you see where?

 

[Dick]

You got the correct answer somehow, but your working is flawed. Can you see where?

Ooops. Yeah, I agree. I somehow just read through the bad part. -1/0 should have been a tip off.

 

[bllnsr]

oh yes -1/0 is -∞ but I made it zero
a/3 +1 = -1/2x
putting x = 0
a/3 +1 = -1/2(0)
a/3 +1 = -1/0
-1/0 is -∞
a/3 +1 = -∞
if a = -3 is the correct answer how to get this value

 

[skiller]

The question states "the coefficient of the term in x is zero".

What do you think this coefficient is?

 

[bllnsr]

somebody told me this general formula
T_{r+1} = \binom{n}{r}a^n b^r
will be used to find 'a' and the statement "the coefficient of the term in x is zero" means
that \binom{n}{r} is 0 and what I did previously is wrong.
I have math exam tomorrow and this is the only question that I cannot solve

 

[Dick]

somebody told me this general formula
T_{r+1} = \binom{n}{r}a^n b^r
will be used to find 'a' and the statement "the coefficient of the term in x is zero" means
that \binom{n}{r} is 0 and what I did previously is wrong.
I have math exam tomorrow and this is the only question that I cannot solve

What you did before is almost right. When you get to 1+(2ax/3)+2x=1+(2a/3+2)x the part you want to make 0 is just the coefficient of x, (2a/3+2). Ignore the 1, it doesn't have anything to do with x.

 

[bllnsr]

@Dick
can you please show me last two steps of how to solve it for a

 

[Dick]

@Dick
can you please show me last two steps of how to solve it for a

Ok, just for you. 2a/3+2=0, 2a/3=(-2) (subtract 2 from both sides), 2a=(-2)*3 (multiply both sides by 3), a=(-2)*3/2=(-3) (divide both sides by 2).

 

[bllnsr]

Thanks

 

[Dick]

Thanks

You're welcome. Notice no 1/0 appears. If it does that's a pretty sure sign something is wrong.