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

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Answer 1 · 2018-03-13T12:28:52

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

# #
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#

# #

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#

# #

Simplify to get:

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

# #

That's it!