Question

How do you expand `(2y+1)^6`?

Answer 1

See a solution process below:

Explanation:

The coefficients from Pascal's Triangle are:

The triangle values for the exponent 3 are:

`color(red)(1)color(white)(.........)color(red)(6)color(white)(.........)color(red)(15)color(white)(.........)color(red)(20)color(white)(.........)color(red)(15)color(white)(.........)color(red)(6)color(white)(.........)color(red)(1)`

Therefore `(color(blue)(2y) + color(green)(1))^6` can be multiplied as:

`color(red)(1)(color(green)(1)^0(color(blue)(2y))^6) + color(red)(6)(color(green)(1)^1(color(blue)(2y))^5) + color(red)(15)(color(green)(1)^2(color(blue)(2y))^4) + color(red)(20)(color(green)(1)^3(color(blue)(2y))^3) + color(red)(15)(color(green)(1)^4(color(blue)(2y))^2) + color(red)(6)(color(green)(1)^5(color(blue)(2y))^1) + color(red)(1)(color(green)(1)^6(color(blue)(2y))^0)`

`color(red)(1)(color(green)(1) * color(blue)(64y^6)) + color(red)(6)(color(green)(1) * color(blue)(32y^5)) + color(red)(15)(color(green)(1) * color(blue)(16y^4)) + color(red)(20)(color(green)(1) * color(blue)(8y^3)) + color(red)(15)(color(green)(1) * color(blue)(4y^2)) + color(red)(6)(color(green)(1) * color(blue)(2y)) + color(red)(1)(color(green)(1) * color(blue)(1))`

`64y^6 + 192y^5 + 240y^4 + 160y^3 + 60y^2 + 12y + 1`