Question

Write the sum of the first three terms in the binomial expansion, expressing the result in simplified form.(x – 4y)^7

Answer 1

`(x-4y)^7=x^7-28x^6y+336x^5y^2\ldots`

Explanation:

We have the following expression:

`\mleft(x-4y\mright)^7`

In this case we can apply the binomial theorem, which is the following:

`(a+b)^n=\sum ^n_(i\mathop=0)((n!)/(i!(n-i)!)a^(n-i)\cdot b^i`

we replace and calculate for the first three terms:

`\begin{gathered} 1st=\sum ^7_{i\mathop{=}0}((7!)/(0!(7-0)!)x^(7-0)\cdot(-4y)^0=1\cdot x^7\cdot1=x^7 \\ 2nd=\sum ^7_{i\mathop{=}1}((7!)/(1!(7-1)!)x^(7-1)\cdot(-4y)^1=7\cdot x^6\cdot-4y^1=-28x^6y \\ 3rd=\sum ^7_{i\mathop{=}2}((7!)/(2!(7-2)!)x^(7-2)\cdot(-4y)^2=21\cdot x^5\cdot16y^2=336x^5y^2 \end{gathered}`