Find the 9th term in the expansion of (3x-y÷3)^12?

1 Answer
Mar 13, 2018

#T_{9}\ =\ \frac{55x^4y^8}{9}#
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Explanation:

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The #\ \ (r+1)\ th\ \ # term of an expression #\ \ (a+x)^n\ \ # by using binomial theorem can be found by the general term formula:

#T_{r+1}\ =\ ((n),(r))a^{n-r}x^r#

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Here, we are required to find the #\ \ 9th\ \ # term, so put #\ \ r=8\ \ # in the above stated formula to get:

#T_{8+1}\ =\ ((12),(8))(3x)^{12-8}(y/3)^8#

#T_{9}\ =\ \frac{12!}{8!(12-8)!}\cdot(3x)^4(y/3)^8#

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Simplify to get:

#T_{9}\ =\ \frac{55x^4y^8}{9}#

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That's it!