Question

Consider the expansion (x + 2)^11 What are the number of terms in this expansion. Find the term containing x^2

Answer 1

Are you familiar with the binomial expansion?

Answer 2

Our teacher briefly explained it and that was that..

Answer 3

The binomial theorem says that:
\[(x+y)^n = \sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right) x^{n-k} y^k\]
So we have n=11, I'll give an example of how to find another term and you can use that to find the term with x^2.
Say we want the term with x^4. Then we want the x^(n-k) to be x^4, so n-k=11-k=4, so k must be 7.
So we find the term with n=11 and k=7. First we have the binomial co-efficient \[\left(\begin{matrix}11 \\ 7\end{matrix}\right)=330 \] (use the nCr button on the calculator)
Then we will have x^(n-k)=x^(11-7)=x^4, and 2^k=2^7.
So when we multiply all these parts together we get 330x^4 times 2^7 = 42,240x^4 as the term containing x^4.
Are you able to apply this to find the term containing x^2?
Please say if you don't understand something I've said or if I can go through the question step by step.

Answer 4

Mm okay. How would you find the numbers in the expansion though?

Answer 5

You have the three components for each term:
-The x component, which will have power n-k.
-The y component (or in this question, y=2 so we will have a power of 2), so for term k we will have 2^k.
-The binomial coefficient.

Since we know the value of y the numbers in the expansion will come from the binomial coefficient multiplied by the respective power of 2. Let me give another example:
Say we want the term with x^7. Then n-k=7, so 11-k=7 and k=4.

So for the numbers which multiply x^7, we will have:
2^k=2^4=16.
The binomial coefficient 11C4=330.
Multiplying these together we get the term 5,280x^7.

Does this make more sense? Can you see where the numbers are coming from?

Answer 6

Yeah, thank you :)

Answer 7

Great :) What did you get as your x^2 term?