Chapter 8.4, Problem 14E

In Exercises 13-44, use mathematical induction to prove that each statement is true for all natural numbers n.

13. $2+4+6+\mathrm{....}+2n=n^2+n$

14. $1+3+5+\mathrm{....}+\left(2n-1\right)=n^2$

15. $4+8+12+\mathrm{....}+4n=2n\left(n+1\right)$

16. $3+6+9+\mathrm{....}+3n=\frac{3n\left(n+1\right)}{2}$

17. $1+5+9+\mathrm{....}+\left(4n-3\right)=n\left(2n-1\right)$

18. $3+8+13+\mathrm{....}+\left(5n-2\right)=\frac{n\left(5n+1\right)}{2}$

19. $3+9+27+\mathrm{....}+3^n=\frac{3\left(3^n-1\right)}{2}$

20. $5+25+125+\mathrm{....}+5^n=\frac{5\left(5^n-1\right)}{4}$

21. $1\cdot 2+3\cdot 4+5\cdot 6+\mathrm{...}+\left(2n-1\right)\left(2n\right)=\frac{1}{3}n\left(n+1\right)\left(4n-1\right)$

22. $1\cdot 3+2\cdot 4+3\cdot 5+\mathrm{...}+\left(n\right)\left(n+2\right)=\frac{1}{6}n\left(n+1\right)\left(2n+7\right)$

23. $\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\mathrm{...}+\frac{1}{n\left(n+1\right)}=\frac{n}{n+1}$

24. $\frac{1}{2\cdot 4}+\frac{1}{4\cdot 6}+\frac{1}{6\cdot 8}+\mathrm{...}+\frac{1}{2n\left(2n+2\right)}=\frac{n}{4\left(n+1\right)}$

25. $2\le 2^n$ 26. $2n+1\le 3^n$

27. $n\left(n+2\right)<{\left(n+1\right)}^2$ 28. $n\le n^2$

29. $\frac{n!}{n}=\left(n-1\right)!$ 30. $\frac{n!}{\left(n+1\right)!}=\frac{1}{n+1}$

31. $1^2+2^2+3^2+\mathrm{...}+n^2=\frac{n\left(n+1\right)(2n+1)}{6}$

32. $1^3+2^3+3^3+\mathrm{...}+n^3=\frac{n^2{\left(n+1\right)}^2}{4}$

33. $2$ is a factor of $n^2+n$ 4, $2$ is a factor of $n^3+5n$

35. $6$ is a factor of $n\left(n+1\right)\left(n+2\right)$ .

36. $3$ is a factor of $n\left(n+1\right)\left(n-1\right)$ .

37. $\left(1+a^n\right)\ge na,a>1$

38. ${\left(1+a\right)}^n\ge 1+na,a>0$

39. ${\displaystyle \underset{i=1}{\overset{n}{\sum }}3\cdot 4^i=4\left(4^n-1\right)}$

40. ${\displaystyle \underset{i=1}{\overset{n}{\sum }}8\cdot 9^i=9\left(9^n-1\right)}$

41. $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{n}\right)=n+1$

42. $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\mathrm{...}+\frac{1}{\sqrt{n}}\ge \sqrt{n}$

43. ${\left(ab\right)}^n=a^n{b}^n$

44. ${\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n}$