What is the coefficient of the x5y5-term in the binomial expansion of (2x 3y)10?

Answer 1

To find the coefficient of the (x^5y^5) term in the binomial expansion of ((2x + 3y)^{10}), we use the binomial theorem. The general term in the expansion is given by (\binom{n}{k} (a)^{n-k} (b)^k). Here, (n = 10), (a = 2x), and (b = 3y). We need (k) such that (n-k = 5) (for (x^5)) and (k = 5) (for (y^5)), thus (k = 5).

Calculating the term: [ \binom{10}{5} (2x)^5 (3y)^5 = \binom{10}{5} \cdot 2^5 \cdot 3^5 \cdot x^5 \cdot y^5. ] Now, (\binom{10}{5} = 252), (2^5 = 32), and (3^5 = 243). Therefore, the coefficient is: [ 252 \cdot 32 \cdot 243 = 196608. ] Thus, the coefficient of the (x^5y^5) term is 196608.