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andre [41]
3 years ago
8

What is the binomial expansion of (x + 2y)7?

Mathematics
2 answers:
nikklg [1K]3 years ago
8 0

Answer: Option 'D' is correct.

Step-by-step explanation:

Since we have given that

(x + 2y)^7

We need to find the binomial expansion :

\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\\\\

So,

\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

So, we get

x^7+14x^6y+84x^5y^2+280x^4y^3+560x^3y^4+672x^2y^5+448xy^6+128y^7

Hence, Option 'D' is correct.

Oduvanchick [21]3 years ago
4 0
<span>x^7+14x^6y+84x^5y^2+280x^4y^3+560x^3y^4+672x^2y^5+448xy^6+128y^7
answer D
</span>
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