Solution. Each follows the same three steps; only step 3 differs.

(a). At \(n=1\): left \(=1\), right \(=\frac {1(2)}{2}=1\). Assume \(1+4+\cdots +(3k-2)=\frac {k(3k-1)}{2}\) and add the next term \(3(k+1)-2=3k+1\): \[\frac {k(3k-1)}{2}+(3k+1)=\frac {3k^{2}-k+6k+2}{2}=\frac {3k^{2}+5k+2}{2} =\frac {(k+1)(3k+2)}{2},\] which is the formula with \(k+1\) in place of \(n\), since \(3(k+1)-1=3k+2\).

(b). At \(n=1\): left \(=1\), right \(=\frac {1(2)(3)}{6}=1\). Assume the formula at \(k\) and add \((k+1)^{2}\): \begin {align*} \frac {k(k+1)(2k+1)}{6}+(k+1)^{2} &=\frac {(k+1)\left [k(2k+1)+6(k+1)\right ]}{6}\\ &=\frac {(k+1)\left (2k^{2}+7k+6\right )}{6} =\frac {(k+1)(k+2)(2k+3)}{6}, \end {align*}

which is the formula at \(k+1\), since \(2(k+1)+1=2k+3\).

(c). At \(n=1\): left \(=1\), right \(=2^{1}-1=1\). Assume \(1+2+\cdots +2^{k-1}=2^{k}-1\) and add \(2^{k}\): \[2^{k}-1+2^{k}=2\cdot 2^{k}-1=2^{k+1}-1 .\]

(d). At \(n=1\): left \(=1+r\), right \(=\frac {1-r^{2}}{1-r}=1+r\) since \(1-r^{2}=(1-r)(1+r)\). Assume the formula at \(k\) and add \(r^{k+1}\): \[\frac {1-r^{k+1}}{1-r}+r^{k+1} =\frac {1-r^{k+1}+r^{k+1}(1-r)}{1-r} =\frac {1-r^{k+2}}{1-r},\] the \(r^{k+1}\) terms cancelling. The condition \(r\neq 1\) is needed throughout, since the right side is undefined at \(r=1\) — where the sum is simply \(n+1\).

Solution. (a). At \(n=1\), \(1+2=3\), divisible by \(3\). Assume \(k^{3}+2k=3m\) for some integer \(m\). Then \begin {align*} (k+1)^{3}+2(k+1)&=k^{3}+3k^{2}+3k+1+2k+2\\ &=\left (k^{3}+2k\right )+3k^{2}+3k+3\\ &=3m+3\left (k^{2}+k+1\right ), \end {align*}

a multiple of \(3\).

(b). At \(n=1\), \(6+4=10\), divisible by \(5\). Assume \(6^{k}+4=5m\). Then \[6^{k+1}+4=6\left (6^{k}+4\right )-20=6(5m)-20=5(6m-4),\] a multiple of \(5\). Subtracting \(20\) is what makes the inductive hypothesis usable: \(6\cdot 4=24\) is four too many, and \(24-20=4\) restores the constant term.

(c). At \(n=1\), \((1+2)!=6>3=3^{1}\). Assume \((k+2)!>3^{k}\). Then \[\left (k+3\right )!=(k+3)\cdot (k+2)!>(k+3)\cdot 3^{k}\geq 4\cdot 3^{k}>3\cdot 3^{k} =3^{k+1},\] using \(k\geq 1\) so that \(k+3\geq 4\). Hence \((n+2)!>3^{n}\) for all \(n\geq 1\).

Solution. (a). At \(n=1\) the left side is \(1\) and the right is \(\frac {1}{3}(1)(3)=1\). Assume the formula at \(n=k\) and add the next term, which is \(\left (2(k+1)-1\right )^{2}=(2k+1)^{2}\): \begin {align*} \frac {1}{3}k\left (4k^{2}-1\right )+(2k+1)^{2} &=\frac {k(2k-1)(2k+1)+3(2k+1)^{2}}{3}\\ &=\frac {(2k+1)\left [k(2k-1)+3(2k+1)\right ]}{3} =\frac {(2k+1)\left (2k^{2}+5k+3\right )}{3}\\ &=\frac {(2k+1)(2k+3)(k+1)}{3} =\frac {1}{3}(k+1)\left (4(k+1)^{2}-1\right ), \end {align*}

since \(4(k+1)^{2}-1=(2k+1)(2k+3)\). That is the formula at \(k+1\).

(b). At \(n=1\), \(8-1=7\), divisible by \(7\). Assume \(8^{k}-1=7m\). Then \[8^{k+1}-1=8\left (8^{k}-1\right )+7=8(7m)+7=7(8m+1),\] a multiple of \(7\).