Question

Use Pascal's triangle and the binomial theoremWhat is the coefficient of the 4th term of (x+y)^6?

Answer 1

`\begin{gathered} \text{(x + y)}^6=\text{ }x^6\text{ + 6}x^5y\text{ + 15}x^4y^2\text{ + 20}x^3y^3\text{ + }15x^2y^4\text{ + 6}x^{}y^5\text{ + }y^6_{} \\ \text{coefficient of the 4th term = 20} \end{gathered}`Step-by-step explanation:
`\begin{gathered} \text{Given:} \\ (x+y)^6 \end{gathered}`

To expand the above, we will use the binomial theorem and pascal's triangle

Using binomial theorem expansion:

`\text{(x + y)}^6=^6C_0x^6\text{ + }^6C_1x^5y^1\text{ + }^6C_2x^4y^2+\text{ }^6C_3x^3y^3\text{ + }^6C_4x^2y^4\text{ + }^6C_5x^1y^5\text{ + }^6C_6x^0^{}y^6`

The values of the combination can be gotten from pascal's triangle. Pascals triangle gives the coefficient of each term

The exponent is 6, we will find the coefficient when the exponent is 6:

`1,\text{ 6, 15, 20, }15,\text{ 6, }1`
`\begin{gathered} \text{(x + y)}^6=^6C_0x^6\text{ + }^6C_1x^5y^1\text{ + }^6C_2x^4y^2+\text{ }^6C_3x^3y^3\text{ + }^6C_4x^2y^4\text{ + }^6C_5x^1y^5\text{ + }^6C_6x^0y^6 \\ ^6C_0\text{ = 1, }^6C_1=\text{ 6 }^6C_2\text{ = 15, }^6C_3\text{ = 20, }^6C_4\text{ = 15, }^6C_5\text{ = 6, }^6C_6\text{ = 1} \\ \\ \text{(x + y)}^6=\text{ 1(}x^6)\text{ + 6(}x^5y^1)\text{ + 15(}x^4y^2)\text{ + 20(}x^3y^3)\text{ + }15(x^2y^4)\text{ + 6(}x^1y^5)\text{ + 1(}x^0y^6_{}) \end{gathered}`
`\begin{gathered} \text{The expanded form:} \\ \text{(x + y)}^6=\text{ }x^6\text{ + 6}x^5y\text{ + 15}x^4y^2\text{ + 20}x^3y^3\text{ + }15x^2y^4\text{ + 6}x^{}y^5\text{ + }y^6_{} \end{gathered}`
`\begin{gathered} \text{ }x^6\text{ + 6}x^5y\text{ + 15}x^4y^2\text{ + 20}x^3y^3\text{ + }15x^2y^4\text{ + 6}x^{}y^5\text{ + }y^6_{} \\ \text{lett 1st term = }x^6 \\ 2nd\text{ term = 6}x^5y\text{ } \\ 3rd\text{ term = 15}x^4y^2\text{ } \\ 4th\text{ term = 20}x^3y^3\text{ } \\ \\ \text{The coefficient of a term is the constant attached to the term} \\ \text{The constant attached to the 4th term is 20} \\ \text{Hence, the coefficient of the 4th term = 20} \end{gathered}`