Find the coefficient of $u^2 v^9$ in the expansion of $(2u - 3v^3 + v^2 - 5u^2 + 1)^5.$
+131 We use the Binomial Theorem to expand the given expression. We have:
(2u - 3v + u^2 - v^2)^9 = ∑(k=0 to 9) C(9,k)(2u)^(9-k)(-3v)^k(u^2 - v^2)^(9-k)
To find the coefficient of u^2 v^9, we need to choose k such that (9-k) powers of (u^2 - v^2) multiply out to u^2 and k powers of (-3v) multiply out to v^9. From the first term (2u)^(9-k), we need to choose two powers of u to multiply out to u^2. Therefore, we choose (9-k-2) = (7-k) of the remaining terms to be (u^2 - v^2), and the remaining k terms to be (-3v). This gives us the equation:
2^(9-k) * (-3)^k * C(9,k) * (u^2 - v^2)^(7-k) * (-3v)^k = u^2 v^9 * coefficient
We need to find the value of the coefficient that satisfies this equation. Plugging in u^2 = v^9 = 1, we get:
2^(9-k) * (-3)^k * C(9,k) * (-3)^(k) = coefficient
Simplifying, we get:
coefficient = 2^(9-k) * 3^(2k) * C(9,k)
To find the coefficient of u^2 v^9, we need to find k such that (9-k-2) = (7-k) powers of (u^2 - v^2) multiply out to u^2, and k powers of (-3v) multiply out to v^9. This means that:
7 - k = 2
k = 5
Therefore, the coefficient of u^2 v^9 is:
2^(9-5) * 3^(2*5) * C(9,5) = 2^4 * 3^10 * 126 = 2,176,782,080
Therefore, the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9 is 2,176,782,080.
+274 I got 1920 @SportzGuy2310 when I expanded it using the binomial theorem.
