Solutions to Try Its

1. a. 35
b. 330

2. a. ${x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}$
b. $8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}$

3. $-10,206{x}^{4}{y}^{5}$

Solutions to Odd-Numbered Exercises

1. A binomial coefficient is an alternative way of denoting the combination $C\left(n,r\right)$. It is defined as $\left(\begin{array}{c}n\\ r\end{array}\right)=C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$.

3. The Binomial Theorem is defined as ${\left(x+y\right)}^{n}=\sum _{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}$ and can be used to expand any binomial.

5. 15

7. 35

9. 10

11. 12,376

13. $64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}$

15. $27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}$

17. $1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}$

19. $1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}$

21. $\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}$

23. ${a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}$

25. ${a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}$

27. $3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}$

29. ${x}^{24}-8{x}^{21}\sqrt{y}+28{x}^{18}y$

31. $-720{x}^{2}{y}^{3}$

33. $220,812,466,875,000{y}^{7}$

35. $35{x}^{3}{y}^{4}$

37. $1,082,565{a}^{3}{b}^{16}$

39. $\frac{1152{y}^{2}}{{x}^{7}}$

41. ${f}_{2}\left(x\right)={x}^{4}+12{x}^{3}$

43. ${f}_{4}\left(x\right)={x}^{4}+12{x}^{3}+54{x}^{2}+108x$

45. $590,625{x}^{5}{y}^{2}$

47. $\left(\begin{array}{c}n\\ k - 1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)=\left(\begin{array}{c}n+1\\ k\end{array}\right)$; Proof:
$\begin{array}{}\\ \\ \\ \left(\begin{array}{c}n\\ k - 1\end{array}\right)+\left(\begin{array}{l}n\\ k\end{array}\right)\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k - 1\right)!\left(n-\left(k - 1\right)\right)!}\\ =\frac{n!}{k!\left(n-k\right)!}+\frac{n!}{\left(k - 1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!}{\left(n-k+1\right)k!\left(n-k\right)!}+\frac{kn!}{k\left(k - 1\right)!\left(n-k+1\right)!}\\ =\frac{\left(n-k+1\right)n!+kn!}{k!\left(n-k+1\right)!}\\ =\frac{\left(n+1\right)n!}{k!\left(\left(n+1\right)-k\right)!}\\ =\frac{\left(n+1\right)!}{k!\left(\left(n+1\right)-k\right)!}\\ =\left(\begin{array}{c}n+1\\ k\end{array}\right)\end{array}$

49. The expression ${\left({x}^{3}+2{y}^{2}-z\right)}^{5}$ cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.