Question

What is the numerical coefficient of the a^4 b^4 term in the expansion of (1/3a^2 - 2b)^6

enter your answer, in the simplest fractional form, in the box​

Answer 1

Use the binomial theorem:

`\left(\frac13a^2-2b\right)^6=\displaystyle\sum_(k=0)^6\binom6k\left(\frac13a^2\right)^(6-k)(-2b)^k=\sum_(k=0)^6\binom6k((-6)^k)/(729)a^(12-2k)b^k`

where

`\dbinom nk=(n!)/(k!(n-k)!)`

is the binomial coefficient.

We get the a⁴ b⁴ term when k = 4, which gives the coefficient

`\dbinom64((-6)^4)/(729)=\boxed{\frac{80}3}`