George's question at Yahoo Answers regarding the binomial theorem

[MarkFL]

Here is the question:

Which term of the expansion of (x^6 + 2 )^{18} contains x^{36}?

Please explain
thank you

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

 

[MarkFL]

Hello George,

By the binomial theorem, we have:

$$\left(x^6+2 \right)^{18}=\sum_{k=0}^{18}\left[{18 \choose k}\left(x^6 \right)^{18-k}2^k \right]=\sum_{k=0}^{18}\left[{18 \choose k}x^{6(18-k)}2^k \right]$$

Hence, the term which contains $x^{36}$ will be the term for which:

$$6(18-k)=36$$

$$18-k=6$$

$$k=12$$

And so, this term is:

$${18 \choose 12}x^{36}2^{12}=18564\cdot x^{36}\cdot 4096=76038144x^{36}$$