Practice 2 Solutions

1. $nCr(15, 7)$

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

6,435

2. $nCr(8, 5)$

${(r+1)}^{th} \mathrm{term}=5+1=6$

56

3. $nCr(14, 10)$

${(r+1)}^{th} \mathrm{term}=10+1=11$

1,001

4. $nCr(10, 3)$

${(r+1)}^{th} \mathrm{term}=3+1=4$

120

5. Determine the sum of row 9.

$2^9=1+9+36+84+126+126+84+36+9+1$

512

6. Write the sum of the ${85}^{th}$ row as a base raised to a power. Do not evaluate.

$n=85$

$2^{85}$

7. ${(x–y)}^7$

$$\begin{gathered} \mathrm{Row} 7: 1, 7, 21, 35, 35, 21, 7, 1 \\ 1x^7{\left(–y\right)}^0+7x^6{\left(–y\right)}^1+21x^5{\left(–y\right)}^2+35x^4{\left(–y\right)}^3+35x^3{\left(–y\right)}^4+21x^2{\left(–y\right)}^5+7x^1{\left(–y\right)}^6+1x^0{\left(–y\right)}^7 \end{gathered}$$

$x^7–7x^{6}y+21x^5{y}^2–35x^4{y}^3+35x^3{y}^4–21x^2{y}^5+7x{y}^6–y^7$

8. ${(m+2)}^4$

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

$m^4+8m^3+24m^2+32m+16$

9. Determine the seventh term of the binomial: ${(b–c)}^{11}$

$$\begin{gathered} r=7–1=6 \\ nCr\left(11, 6\right)=\frac{11!}{6!\left(11–6\right)!}=462 \\ n=11 \\ 11–5=6 \\ 462{\left(b\right)}^5{\left(–c\right)}^6 \\ \begin{array}{}\\ \\ \end{array} \end{gathered}$$

$462 b^5{c}^6$

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

$$\begin{gathered} \mathrm{Middle} \mathrm{term} = 9\mathrm{th} \mathrm{term} \\ r=9–1=8 \\ nCr\left(16, 8\right)=\frac{16!}{8!\left(16–8\right)!}=12,870 \\ 12,870{\left(a\right)}^8·{\left(1\right)}^8 \end{gathered}$$

$12,870a^8$

11. ${(2x–1)}^5$

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

$32x^5–80x^4+80x^3–40x^2+10x–1$

12. ${(a+2b)}^4$

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

$a^4+8a^3{b}^1+24a^2{b}^2+32a{b}^3+16b^4$