The Binomial Theorem for positive integer powers can be written:
(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k
where ((n),(k)) = (n!)/(k! (n-k)!)
Note that some people like to call the first row of Pascal's triangle the 0th. Others like me prefer to call it the 1st.
Counting from 1, the n+1st row of Pascal's triangle consists of the numbers ((n),(0)), ((n),(1)), ... ((n), (n)).
So rather than 'calculate' the individual coefficients for (a+b)^n, you can read them off from the (n+1)st row of Pascal's triangle...
![enter image source here]()
For example, if we were calculating (a+b)^12 then the coefficients would be 1, 12, 66, 220,..., 1.
Typically our a and b are not plain variables, but have a multiplier, e.g. (2a+3b)^n. In such a case you need to multiply the binomial coefficient by a suitable multiple of the powers of (2a) and (3b), e.g. ((n),(k)) (2a)^(n-k) (3b)^k = ((n),(k))2^(n-k)3^k a^(n-k) b^k, etc.
