Practice 1 Solutions

1. Write rows 0 through 6 of Pascal’s Triangle.

2. Explain how to find the next row of Pascal’s Triangle.

Sample: Every row starts and ends with the number one. To find all the other numbers, you add every pair of numbers in the previous row to get the next number for the new row.

3. $nCr(12, 7)$

${(r+1)}^{th} \mathrm{term} = 7+1=8$

792

4. $nCr(9, 8)$

${(r+1)}^{th} \mathrm{term}=8+1=9$

9

5. Determine the sum of row 8.

$2^8=1+8+28+56+70+56+28+8+1$

256

6. Write the sum of the ${39}^{th}$ row as a base raised to a power. Do not evaluate.

$n=39$

$2^{39}$

7. ${(a+h)}^6$

$$\begin{gathered} \mathrm{Row} 6: 1, 6, 15, 20, 15, 6, 1 \\ 1a^6{h}^0+6a^5{h}^1+15a^4{h}^2+20a^3{h}^3+15a^2{h}^4+6a^1{h}^5+1a^0{h}^6 \end{gathered}$$

$a^6+6a^{5}h+15a^4{h}^2+20a^3{h}^3+15a^2{h}^4+6a{h}^5+h^6$

8. ${(2–n)}^3$

$$\begin{gathered} \mathrm{Row} 3:\hspace{0.17em}1, 3, 3, 1 \\ 1\left(2^3\right){\left(–1n\right)}^0+3\left(2^2\right){\left(–1n\right)}^1+3\left(2^1\right){\left(–1n\right)}^2+1\left(2^0\right){\left(–1n\right)}^3 \\ 8{\left(–1n\right)}^0+12{\left(–1n\right)}^1+6{\left(–1n\right)}^2+1{\left(–1n\right)}^3 \end{gathered}$$

$8–12n+6n^2–1n^3$

9. Determine the third term of the binomial: ${(x–y)}^4$

$$\begin{gathered} r=3–1=2 \\ nCr\left(4, 2\right)=\frac{4!}{2!\left(4–2\right)!}=6 \\ 6{\left(x\right)}^2·{\left(–y\right)}^2 \end{gathered}$$

$6x^2{y}^2$

10. Determine the sixth term of the binomial: ${(a+1)}^{16}$

$$\begin{gathered} r=6–1=5 \\ nCr\left(16, 5\right)=\frac{16!}{5!\left(16–5\right)!}=4,368 \\ 4,368 {\left(a\right)}^{11}{\left(1\right)}^5 \end{gathered}$$  

$4,368 a^{11}$

11. ${(3x+2)}^4$

$$\begin{gathered} \mathrm{Row} 4: 1, 4, 6, 4, 1 \\ 1{\left(3x\right)}^4{\left(2\right)}^0+4{\left(3x\right)}^3{\left(2\right)}^1+6{\left(3x\right)}^2{\left(2\right)}^2+4{\left(3x\right)}^1{\left(2\right)}^3+1{\left(3x\right)}^0{\left(2\right)}^4 \\ 1\left(81\right)x^4+4\left(27\right)\left(2\right)x^3+6\left(9\right)\left(4\right)x^2+4\left(3\right)\left(8\right)x+1\left(1\right)16 \end{gathered}$$

$81x^4+216x^3+216x^2+96x+16$

12. ${\left(x–\frac{1}{2}y\right)}^5$

$$\begin{gathered} \mathrm{Row} 5: 1, 5, 10, 10, 5, 1 \\ 1x^5{\left(–\frac{1}{2}y\right)}^0+5x^4{\left(–\frac{1}{2}y\right)}^1+10x^3{\left(–\frac{1}{2}y\right)}^2+10x^2{\left(–\frac{1}{2}y\right)}^3+5x^1{\left(–\frac{1}{2}y\right)}^4+1x^0{\left(–\frac{1}{2}y\right)}^5 \\ {x}^5+5x^4\left(–\frac{1}{2}y\right)+10x^3\left(\frac{1}{4}{y}^2\right)+10x^2\left(–\frac{1}{8}{y}^3\right)+5x^1\left(\frac{1}{16}{y}^4\right)+\left(–\frac{1}{32}{y}^5\right) \end{gathered}$$

$x^5–\frac{5}{2}{x}^{4}y+\frac{5}{2}{x}^3{y}^2–\frac{5}{4}{x}^2{y}^3+\frac{5}{16}x{y}^4–\frac{1}{32}{y}^5$