# How do you expand (1+x^3)^4 using Pascal’s Triangle?

Feb 8, 2016

Since there are (4 + 1) = 5 terms in this expansion, we must find the numbers located in the ${5}^{t h}$ term of the Pascal's Triangle. To find the number of terms in an expansion, always add 1 to the exponent, as to include the ${0}^{t h}$ term.

#### Explanation:

Draw a diagram to represent Pascal's Triangle. Each row is the sum of the numbers above it, with 1 at the first row, (1 and 1) at the second row, (1, 2 and 1) in the third row. The following diagram is of Pascal's Triangle:

Counting up from the row with a single 1, we find that row 5 contains the numbers 1, 4, 6, 4 and 1.

To expand, the exponents on the 1 will start at 4 and will decrease until 0. The exponents on the ${x}^{3}$ will increase from 0 to 4. As you can see, in each term the exponents must add up to the expression's exponent, which in this case is 4.

$1 {\left(1\right)}^{4} {\left({x}^{3}\right)}^{0} + 4 {\left(1\right)}^{3} {\left({x}^{3}\right)}^{1} + 6 {\left(1\right)}^{2} {\left({x}^{3}\right)}^{2} + 4 {\left(1\right)}^{1} {\left({x}^{3}\right)}^{3} + 1 {\left(1\right)}^{0} {\left({x}^{3}\right)}^{4}$

Simplifying by using exponent laws:

$1 + 4 {x}^{3} + 6 {x}^{6} + 4 {x}^{9} + {x}^{12}$

When fully expanded, ${\left(1 + {x}^{3}\right)}^{4}$ = $1 + 4 {x}^{3} + 6 {x}^{6} + 4 {x}^{9} + {x}^{12}$. As you can see, in each t

Practice Exercises:

1. Expand ${\left(2 x - 3 y\right)}^{5}$ using Pascal's Triangle.

2. Find the 3rd term in ${\left(x + 3\right)}^{7}$. Hint: Think of finding the appropriate number in the Pascal's Triangle and plugging it in for nCr in ${t}_{r + 1} = n C r {\left(a\right)}^{n - r} \times {b}^{r}$.

Good luck!