What are the first 4 non-zero terms in the binomial expansion of (2+3x)^6

Well in a binomial expansion we need to remember the formula (1+x)^n = 1 + nx + (n(n-1)(x^2))/2! + (n(n-1)(n-2)x^3)/3! + ... (I can draw this out properly on the whiteboard app)There is another formula but that one only works when the exponent is a positive integer. This formula works no matter what the exponent is I suggest we just remember the one formula as it works in all situations :)But this formula requires that our function begins with 1 + ... when ours begins with 2 + ... , we need to take a factor of 2 out of our function.( 2 + 3x )^6 = ( 2 ( 1 + (3/2)x))^6 ( again this would look a lot less awful on the whiteboard)Then we can start filling the formula out where every instance of x is replaced with (3/2) and every instance of n is replaced by 6(I would then fill it out and begin to find the simplest forms of all the fractions up to and including the x^3 term)