Respuesta :
x^7+14x^6y+84x^5y^2+280x^4y^3+560x^3y^4+672x^2y^5+448xy^6+128y^7
answer D
answer D
Answer: Option 'D' is correct.
Step-by-step explanation:
Since we have given that
[tex](x + 2y)^7[/tex]
We need to find the binomial expansion :
[tex]\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i\\\\a=x,\:\:b=2y\\\\=\sum _{i=0}^7\binom{7}{i}x^{\left(7-i\right)}\left(2y\right)^i\\\\[/tex]
So,
[tex]\frac{7!}{0!\left(7-0\right)!}x^7\left(2y\right)^0+\frac{7!}{0!\left(7-0\right)!}x^7\left(2y\right)+\frac{7!}{2!\left(7-2\right)!}x^5\left(2y\right)^2+\frac{7!}{3!\left(7-3\right)!}x^4\left(2y\right)^3+\frac{7!}{4!\left(7-4\right)!}x^3\left(2y\right)^4+\frac{7!}{5!\left(7-5\right)!}x^2\left(2y\right)^5+\frac{7!}{6!\left(7-6\right)!}x^1\left(2y\right)^6+\frac{7!}{7!\left(7-7\right)!}x^0\left(2y\right)^7[/tex]
So, we get
[tex]x^7+14x^6y+84x^5y^2+280x^4y^3+560x^3y^4+672x^2y^5+448xy^6+128y^7[/tex]
Hence, Option 'D' is correct.
