Respuesta :
Answer:
d. [tex]-15x^3y^7[/tex]
Step-by-step explanation:
By the binomial theorem,
[tex](p+q)^n=\sum_{r=0}^{n} ^nC_r (p)^{n-r} q^r[/tex]
Where,
[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]
Thus,
[tex](\frac{1}{2}x-y)^{10}=(\frac{x}{2}+(-y))^{10})=\sum_{r=0}^{n} ^10C_r (\frac{x}{2})^{10-r} (-y)^{r}[/tex]
For the 8th term, r = 7,
Thus, the 8th term would be,
[tex]^{10}C_7 (\frac{x}{2})^{10-7} (-y)^{7}[/tex]
[tex]=\frac{10!}{7!(10-7)!} (\frac{x}{2})^3 (-y)^7[/tex]
[tex]=-\frac{10\times 9\times 8}{3\times 2\times 1}\times \frac{x^3}{8}\times y^7[/tex]
[tex]=-\frac{720x^3y^7}{48}[/tex]
[tex]=-15x^3y^7[/tex]
Hence, option 'D' is correct.
