Question

Find the coefficient a of the given term in the expansion of of the binomial. Binomial: (x^2+3)^12 Term: ax^10

Answer 1

We will use the binomial theorem:
http://en.wikipedia.org/wiki/Binomial_theorem
$$
\large
(x^2+3)=\sum_{i=0}^{12} \binom{12}{2}(x^2)^i\times3^{12-i}
$$
From which it it is clear that the \(x^{10^{th}}\) term is
$$
\large
\binom{12}{5}(x^2)^5\times3^7
$$

That's it!

Does this make sense?

Answer 2

I'm pretty confused on this, I just don't understand it @ybarrap

Answer 3

Have you worked with the binomial theorem before?

Answer 4

No this is my first time @ybarrap

Answer 5

Do you know what this means?
$$
\large
(x^2+3)=\sum_{i=0}^{12} \binom{12}{2}(x^2)^i\times3^{12-i}
$$

Answer 6

Yes @ybarrap

Answer 7

which by the way should actually be
$$
\large
(x^2+3)=\sum_{i=0}^{12} \binom{12}{\color{red}{i}}(x^2)^i\times3^{12-i}
$$

Answer 8

Ok. So if you look at that carefully, look at the \(x^2\) term. What \(i\) would make \((x^2)^i\) equal to \(x^{10}\)?

Answer 9

Yes

Answer 10

What is the value of \(\large i\) that makes
$$
\Large
(x^2)^i=x^{10}
$$

Answer 11

$$
\Large
(x^2)^i=x^{2i}
$$
Does this help?

Answer 12

Yes

Answer 13

Well then what is \(i\)?

Answer 14

$$
\huge{
x^{2i}=x^{10}
}
$$
What is \(\Huge i\)?

Answer 15

$$
\Huge
2i=10
$$
What is \(\huge i\)?

Answer 16

Would i be 5?

Answer 17

Is \(2\times 5=10\)?

Answer 18

Oh ok I didn;t know you were suppose to multiply

Answer 19

The point is that you need to know which of the terms in the sum are the ones associated with \(x^{10}\) and i=5 is the key.

Knowing that i=5, then you can just take out of the sum (in the binomial series) the constants associated with i=5:
$$
\large{
(x^2+3)=\sum_{i=0}^{12} \binom{12}{\color{red}{i}}(x^2)^i\times3^{12-i}\\
=\binom{12}{0}(x^2)^0\times3^{12-0}+\binom{12}{1}(x^2)^1\times3^{12-1}+\cdots+\\\color{red}{\binom{12}{5}(x^2)^{5}\times3^{12-5}}+\cdots+\binom{12}{12}(x^2)^{12}\times3^{12-12}
}
$$.
The term in red is the one associated with \(x^{10}\)

So now all you have to do is calculate:
$$
\large
\binom{12}{5}\times3^7
$$
Which is the \(a\) in \(ax^{10}\).

I hope this makes sense.

Answer 20

It will thank you

Answer 21

What do you get when you calculate? I'll verify for you.

Answer 22

Use this if you need help - http://www.wolframalpha.com/input/?i=%28x^2%2B3%29^12

Answer 23

I got 1732104 I think I calculated it wrong

Answer 24

From which you can validate that our solution is correct - http://www.wolframalpha.com/input/?i=%2812+choose+5%29*3^7