Question

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

A: 2x7 + 14x6y + 42x5y2 + 70x4y3 + 70x3y4 + 42x2y5 + 14xy6 + 2y7
B: x7 + 14x6y + 42x5y2 + 70x4y3 + 70x3y4 + 42x2y5 + 14xy6 + y7
C: x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7
D: x7 + 14x6y + 84x5y2 + 280x4y3 + 560x3y4 + 672x2y5 + 448xy6 + 128y7

Answer 1

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

Answer 2

Answer: Option 'D' is correct.

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,

`(7!)/(0!\left(7-0\right)!)x^7\left(2y\right)^0+(7!)/(0!\left(7-0\right)!)x^7\left(2y\right)+(7!)/(2!\left(7-2\right)!)x^5\left(2y\right)^2+(7!)/(3!\left(7-3\right)!)x^4\left(2y\right)^3+(7!)/(4!\left(7-4\right)!)x^3\left(2y\right)^4+(7!)/(5!\left(7-5\right)!)x^2\left(2y\right)^5+(7!)/(6!\left(7-6\right)!)x^1\left(2y\right)^6+(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.