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Abigail's question at Yahoo! Answers regarding binomial expansion

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MarkFL

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Feb 24, 2012
13,775
Here is the original question:

What is the third term of the expansion of (2x+y^2)^9?

I don't understand how to solve this problem without just working the entire thing out! If you could explain how to do it that would be great! Thank you so much, any help would be much appreciated!
Here is a link to the original question:

What is the third term of the expansion of (2x+y^2)^9? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

Administrator
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Feb 24, 2012
13,775
Hello Abigail,

The binomial theorem gives us:

$\displaystyle (a+b)^n=\sum_{k=0}^n{n \choose k}a^{n-k}b^k$

and so:

$\displaystyle (2x+y^2)^9=\sum_{k=0}^9{9 \choose k}(2x)^{9-k}(y^2)^k$

Now, the third term corresponds to $\displaystyle k=2$, hence this term is:

$\displaystyle {9 \choose 2}(2x)^{9-2}(y^2)^2=36\cdot(2x)^7y^4=4608x^7y^4$
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Abigail!

$\text{What is the third term of the expansion of }\,(2x+y^2)^9\,?$
Recall that $n=9$ on Pascal's Triangle gives:.$1,\;9,\;36,\;84,\;126,\;126,\;84,\;36,\;9,\;1$

So that $(a+b)^9$ begins with: .$a^9 + 9a^8b + 36a^7b^2 + 84a^6b^3 + \cdots$

The third term is: .$36(2x)^7(y^2)^2 \:=\:36(128x^7)(y^4) \:=\:4608x^7y^4$