In Exercises proofindfirst - proofindlast, prove each assertion using the Principle of Mathematical Induction.
\(\displaystyle { \sum _{j=1}^{n} j^2 = \frac {n(n+1)(2n+1)}{6}}\)
Let \(P(n)\) be the sentence \(\displaystyle { \sum _{j=1}^{n} j^2 = \frac {n(n+1)(2n+1)}{6}}\). For the base case, \(n=1\), we get
\[ \begin{array}{rcl}
\displaystyle { \sum _{j=1}^{1} j^2} & \stackrel {?}{=} & \frac {(1)(1+1)(2(1)+1)}{6} \\ 1^2 & = & 1 \, \checkmark \\ \end{array} \]
We now assume \(P(k)\) is true and use it to show \(P(k+1)\) is true. We have
\[ \begin{array}{rcl}
\displaystyle { \sum _{j=1}^{k+1} j^2} & \stackrel {?}{=} & \frac {(k+1)((k+1)+1)(2(k+1)+1)}{6} \\ \displaystyle { \sum _{j=1}^{k} j^2} + (k+1)^2 & \stackrel {?}{=} & \frac {(k+1)(k+2)(2k+3)}{6} \\ \underbrace {\frac {k(k+1)(2k+1)}{6}}_{\text {Using $P(k)$}} + (k+1)^2 & \stackrel {?}{=} & \frac {(k+1)(k+2)(2k+3)}{6} \\
&& \\
\frac {k(k+1)(2k+1)}{6} + \frac {6(k+1)^2}{6} & \stackrel {?}{=} & \frac {(k+1)(k+2)(2k+3)}{6} \\ \frac {k(k+1)(2k+1)+6(k+1)^2}{6} & \stackrel {?}{=} & \frac {(k+1)(k+2)(2k+3)}{6} \\ \frac {(k+1)(k(2k+1)+6(k+1))}{6} & \stackrel {?}{=} & \frac {(k+1)(k+2)(2k+3)}{6} \\ \frac {(k+1)\left (2k^2+7k+6\right )}{6} & \stackrel {?}{=} & \frac {(k+1)(k+2)(2k+3)}{6} \\ \frac {(k+1)(k+2)(2k+3)}{6} & = & \frac {(k+1)(k+2)(2k+3)}{6} \, \checkmark \\ \end{array} \]
By induction, \(\displaystyle { \sum _{j=1}^{n} j^2 = \frac {n(n+1)(2n+1)}{6}}\) is true for all natural numbers \(n \geq 1\).
\(\displaystyle { \sum _{j=1}^{n} j^3 = \frac {n^2(n+1)^2}{4}}\)
\(2^{n} > 500 n\) for \(n > 12\)
\(3^{n} \geq n^3\) for \(n \geq 4\)
Let \(P(n)\) be the sentence \(3^n > n^3\). Our base case is \(n=4\) and we check \(3^4 = 81\) and \(4^3 = 64\) so that \(3^4 > 4^3\) as required. We now assume \(P(k)\) is true, that is \(3^k > k^3\), and try to
show \(P(k+1)\) is true. We note that \(3^{k+1} = 3 \cdot 3^{k} > 3k^3\) and so we are done if we can show \(3k^3 > (k+1)^3\) for \(k \geq 4\). We can solve the inequality \(3x^3 > (x+1)^3\) using the
techniques of Section RootRadicalFunctions, and doing so gives us \(x > \frac {1}{\sqrt [3]{3}-1} \approx 2.26.\) Hence, for \(k \geq 4\), \(3^{k+1} = 3 \cdot 3^{k} > 3k^3 > (k+1)^3\) so that \(3^{k+1} > (k+1)^3\). By induction, \(3^n > n^3\) is true for all natural numbers \(n \geq 4\).
Use the Product Rule for Absolute Value to show \(\left |x^{n}\right | = |x|^{n}\) for all real numbers \(x\) and all natural numbers \(n \geq 1\)
Let \(P(n)\) be the sentence \(\log \left (x^n \right ) = n \log (x)\). For the duration of this argument, we assume \(x > 0\). The base case \(P(1)\) amounts checking that \(\log \left (x^1\right ) = 1 \log (x)\) which is clearly
true. Next we assume \(P(k)\) is true, that is \(\log \left (x^{k}\right ) = k \log (x)\) and try to show \(P(k+1)\) is true. Using the Product Rule for Logarithms along with the induction
hypothesis, we get
\[\log \left (x^{k+1}\right ) = \log \left (x^{k} \cdot x\right ) = \log \left (x^{k}\right ) + \log (x) = k \log (x) + \log (x) = (k+1) \log (x) \]
Hence, \(\log \left (x^{k+1}\right ) = (k+1) \log (x)\). By induction \(\log \left (x^n \right ) = n \log (x)\) is true for all \(x>0\) and all natural numbers \(n \geq 1\).
Use the Product Rule for Logarithms to show \(\log \left (x^{n}\right ) = n \log (x)\) for all real numbers \(x > 0\) and all natural numbers \(n \geq 1\).