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Answer:
-960
Step-by-step explanation:
1) Expand [tex](x - 2y)^{10}[/tex] using the binomial theorem.
Binomial Theorem: [tex]\left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i[/tex]
1.1) Substitute the values into the theorem.
[tex]\sum _{i=0}^{10}\binom{10}{i}x^{\left(10-i\right)}\left(-2y\right)^i[/tex]
1.2) Expand summation.
[tex]=\frac{10!}{0!\left(10-0\right)!}x^{10}\left(-2y\right)^0+\frac{10!}{1!\left(10-1\right)!}x^9\left(-2y\right)^1+\frac{10!}{2!\left(10-2\right)!}x^8\left(-2y\right)^2+\frac{10!}{3!\left(10-3\right)!}[/tex]
[tex]x^7\left(-2y\right)^3+\frac{10!}{4!\left(10-4\right)!}x^6\left(-2y\right)^4+\frac{10!}{5!\left(10-5\right)!}x^5\left(-2y\right)^5+\frac{10!}{6!\left(10-6\right)!}x^4\left(-2y\right)^6+\frac{10!}{7!\left(10-7\right)!}x^3\left(-2y\right)^7+\frac{10!}{8!\left(10-8\right)!}x^2\left(-2y\right)^8+\frac{10!}{9!\left(10-9\right)!}x^1\left(-2y\right)^9+\frac{10!}{10!\left(10-10\right)!}x^0\left(-2y\right)^{10}[/tex]
1.3) Simplify them.
1.4) You will get:
[tex]=x^{10}-20x^9y+180x^8y^2-960x^7y^3+3360x^6y^4-8064x^5y^5+13440x^4y^6-15360x^3y^7+11520x^2y^8-5120xy^9+1024y^{10}[/tex]
2) We are told to find the coefficient of [tex]x^{7} y^{3}[/tex]. Find it from the simplified expansion. The coefficient is -960.