Find the coefficient of x4 in the expansion of (3x - 2)6
Hello, maptorren!
I must assume you're familiar with the Binomial Expansion.
The term we seek is: .$\displaystyle {6\choose4}(3x)^4(\text{-}2)^2 \;=\;15(81x^4)(4) \;=\;4860x^4$Find the coefficient of $\displaystyle x^4$ in the expansion of $\displaystyle (3x-2)^6$
The coefficient is 4860.
Using the factorial form of the binomial theorem, you want to find the 3rd term of $\displaystyle (3x-2)^6$.
$\displaystyle (3x-2)^6=\sum_{k=0}^{6} \frac{6!}{(6-k)!k!}\ \ (3x)^{6-k}(-2)^k$
Find the 3rd term when k=2,
$\displaystyle \frac{6!}{(6-2)!2!}\ \ (3x)^{6-2}(-2)^2=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{4\cdot 3\cdot2\cdot1\cdot2\cdot1}(3x)^4(4)=15\cdot81x^4\c dot4=4860x^4$