Question

Resultado y explicación por favor.

Answer 1

Respuesta:    $\quad -y^{10}+5y^8-10y^6+10y^4-5y^2+1$

Explicación paso a paso:

$\left(1-y^2\right)^5$

$\mathrm{Aplicar\:el\:teorema\:del\:binomio}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=1,\:\:b=-y^2$

$=\sum _{i=0}^5\binom{5}{i}\cdot \:1^{\left(5-i\right)}\left(-y^2\right)^i$

$\mathrm{Expandir\:sumatorio}:$

$\frac{5!}{0!\left(5-0\right)!}\cdot \:1^5\left(-y^2\right)^0+\frac{5!}{1!\left(5-1\right)!}\cdot \:1^4\left(-y^2\right)^1+\frac{5!}{2!\left(5-2\right)!}\cdot \:1^3\left(-y^2\right)^2+\frac{5!}{3!\left(5-3\right)!}\cdot \:1^2\left(-y^2\right)^3$ $+\frac{5!}{4!\left(5-4\right)!}\cdot \:1^1\left(-y^2\right)^4+\frac{5!}{5!\left(5-5\right)!}\cdot \:1^0\left(-y^2\right)^5$

$\frac{5!}{0!\left(5-0\right)!}\cdot \:1^5\left(-y^2\right)^0=1$

$\mathrm{Simplificar}\:\frac{5!}{1!\left(5-1\right)!}\cdot \:1^4\left(-y^2\right)^1:\quad -5y^2$

$\mathrm{Simplificar}\:\frac{5!}{2!\left(5-2\right)!}\cdot \:1^3\left(-y^2\right)^2:\quad 10y^4$

$\mathrm{Simplificar}\:\frac{5!}{3!\left(5-3\right)!}\cdot \:1^2\left(-y^2\right)^3:\quad -10y^6$

$\mathrm{Simplificar}\:\frac{5!}{4!\left(5-4\right)!}\cdot \:1^1\left(-y^2\right)^4:\quad 5y^8$

$\mathrm{Simplificar}\:\frac{5!}{5!\left(5-5\right)!}\cdot \:1^0\left(-y^2\right)^5:\quad -y^{10}$

$=1-5y^2+10y^4-10y^6+5y^8-y^{10}$

$\mathrm{Reescribir\:en\:la\:forma\:estandar}$

$=-y^{10}+5y^8-10y^6+10y^4-5y^2+1$