Solutions for Binomial Theorem

Solutions to Try Its

1. a. 35
b. 330

2. a. [latex]{x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}[/latex]
b. [latex]8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}[/latex]

3. [latex]-10,206{x}^{4}{y}^{5}[/latex]

Solutions to Odd-Numbered Exercises

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

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

5. 15

7. 35

9. 10

11. 12,376

13. [latex]64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}[/latex]

15. [latex]27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}[/latex]

17. [latex]1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}[/latex]

19. [latex]1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}[/latex]

21. [latex]\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}[/latex]

23. [latex]{a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}[/latex]

25. [latex]{a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}[/latex]

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

29. [latex]{x}^{24}-8{x}^{21}\sqrt{y}+28{x}^{18}y[/latex]

31. [latex]-720{x}^{2}{y}^{3}[/latex]

33. [latex]220,812,466,875,000{y}^{7}[/latex]

35. [latex]35{x}^{3}{y}^{4}[/latex]

37. [latex]1,082,565{a}^{3}{b}^{16}[/latex]

39. [latex]\frac{1152{y}^{2}}{{x}^{7}}[/latex]

41. [latex]{f}_{2}\left(x\right)={x}^{4}+12{x}^{3}[/latex]
Graph of the function f_2.

43. [latex]{f}_{4}\left(x\right)={x}^{4}+12{x}^{3}+54{x}^{2}+108x[/latex]
Graph of the function f_4.

45. [latex]590,625{x}^{5}{y}^{2}[/latex]

47. [latex]\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)[/latex]; Proof:
[latex]\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}[/latex]

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