Jeffrrey's question at Yahoo Answers regarding binomial expansion

[MarkFL]

Here is the question:

Can someone help me please?


Question: The term independent of x in the expansion of ((3x^2) - (t/x))^6 is 2160.

Given that t > 0, find the value of t.


Solution so far:

Formula - (n r) ((x)^n-r) ((y)^r)

So:

(6 r) ((3x^2)^6-r) (-t/x)^r


However, I don't know where to go from here. I am supposed to get:

x^12-3r

and then from that I am supposed to get r = 4

and then t = 2

I have posted a link there to this topic so the OP can view my work.

 

[MarkFL]

Hello Jeffrey,

We are given the expression:

\(\displaystyle \left(3x^2-\frac{t}{x} \right)^6\)

Using the binomial theorem, we may write:

\(\displaystyle \left(3x^2+\left(-\frac{t}{x} \right) \right)^6=\sum_{k=0}^6\left[{6 \choose k}\left(3x^2 \right)^{6-k}\left(-\frac{t}{x} \right)^k \right]\)

We may rewrite this as:

\(\displaystyle \left(3x^2+\left(-\frac{t}{x} \right) \right)^6=\sum_{k=0}^6\left[{6 \choose k}3^{6-k}(-t)^kx^{3(4-k)} \right]\)

Now, in order for a term to be independent of $x$, we require the exponent on $x$ to be zero, which occurs for:

\(\displaystyle 3(4-k)=0\implies k=4\)

We are told this term is equal to $2160$, hence we have:

\(\displaystyle {6 \choose 4}3^{6-4}(-t)^4x^{3(4-4)}=2160\)

\(\displaystyle \frac{6!}{4!(6-4)!}\cdot3^2\cdot t^4=2160\)

\(\displaystyle 15\cdot9\cdot t^4=2160\)

\(\displaystyle t^4=14=2^4\)

For real $0<t$ we then find:

\(\displaystyle t=2\)