Proof of (a). Step 1. For \(n=1\) the left side is \(2(1)-1=1\) and the right side is \(1^{2}=1\). They agree, so the statement is true for \(n=1\).

Step 2. Assume it holds for \(n=k\): \[1+3+5+\cdots +(2k-1)=k^{2}.\]

Step 3. We show it then holds for \(n=k+1\), that is \[1+3+5+\cdots +(2k-1)+\left (2(k+1)-1\right )=(k+1)^{2}.\] Taking the left side and replacing the first \(k\) terms using the inductive hypothesis, \begin {align*} k^{2}+\left (2(k+1)-1\right ) &= k^{2}+2k+2-1\\ &= k^{2}+2k+1\\ &= (k+1)^{2}, \end {align*}

which is the right side. So the statement holds for \(n=k+1\), and by induction it holds for every positive integer \(n\). □

Proof of (b). Step 1. For \(n=1\) the left side is \(4\) and the right side is \(2(1)(1+1)=4\). True for \(n=1\).

Step 2. Assume \(4+8+12+\cdots +4k=2k(k+1)\).

Step 3. Adding the next term \(4(k+1)\) to both sides, \begin {align*} \underbrace {4+8+\cdots +4k}_{=\,2k(k+1)}+4(k+1) &= 2k(k+1)+4(k+1)\\ &= 2(k+1)(k+2), \end {align*}

taking out the common factor \(2(k+1)\). This is the required formula with \(k+1\) in place of \(n\), so the result follows by induction. □

Proof of (c). Step 1. For \(n=1\), the left side is \(\frac {1}{2}\) and the right side is \(1-\frac {1}{2}=\frac {1}{2}\). True.

Step 2. Assume \(\frac {1}{2}+\frac {1}{2^{2}}+\cdots +\frac {1}{2^{k}}=1-\frac {1}{2^{k}}\).

Step 3. Adding \(\frac {1}{2^{k+1}}\) to both sides, \begin {align*} 1-\frac {1}{2^{k}}+\frac {1}{2^{k+1}} &= \frac {2\cdot 2^{k}-2+1}{2\cdot 2^{k}}\\ &= \frac {2\cdot 2^{k}-1}{2\cdot 2^{k}}\\ &= 1-\frac {1}{2\cdot 2^{k}} = 1-\frac {1}{2^{k+1}}, \end {align*}

as required, so the result holds for all \(n\) by induction. □

Proof of (d). Step 1. For \(n=1\), \(x^{1}-y^{1}=x-y\), which is divisible by \(x-y\).

Step 2. Assume \(x^{k}-y^{k}\) is divisible by \(x-y\).

Step 3. Write \(x^{k+1}-y^{k+1}\) by adding and subtracting \(xy^{k}\): \[x^{k+1}-y^{k+1}=x^{k}x-xy^{k}+xy^{k}-y^{k}y =x\left (x^{k}-y^{k}\right )+y^{k}(x-y).\] The first term is divisible by \(x-y\) by the inductive hypothesis; the second has \(x-y\) as an explicit factor. A sum of two multiples of \(x-y\) is again a multiple of \(x-y\), so \(x^{k+1}-y^{k+1}\) is divisible by \(x-y\) and the result follows. □