Let $P(n)$ be a [[Proposition|Propositional Statement]]. To prove [[Proof|prove]] that $P(n)$ is true [[Universal Quantifier|For All]] positive [[Integer|Integers]] $n$, we complete two steps: - *Basis Step*: Verify that $P(1)$ is [[Truth Value|true]]. - *Induction Step*: Let $k$ be an arbitrary positive [[Integer]], prove the [[Conditional]] $P(k) \to P(k+1)$ $\huge {\color{pink}P}(n) =\mat{ {\color{pink}P}(1)\to {\color{pink}P}(k)\to {\color{pink}P} (k+1) } $ >[!example] >Example: Prove that: >$ >1^3+2^3+\dots+n^2= \pa{\frac{n(n+1)}{2}}^2 >$ > >>[!check]- Solution >>The base step is satisfied: >>$ 1^3 = \pa{\frac{1(1+1)}{2}}^2=1 $ >> >> >> >>$ >>1^3+2^3+\dots+k^3= \pa{\frac{k(k+1)}{2}}^2 >>$ >> >>$ >>1^3+2^3+\dots+(k+1)^3= \pa{\frac{(k+1)(k+2)}{2}}^2 >>$ >> >>Using the value of $k$: >>$\begin{align} >>1^3+2^3+\dots+(k+1)^3 >>&= \pa{\frac{k(k+1)}{2}}^2+(k+1)^3\\ >>&= >>\frac{ >>k^2(k+1)^2+4(k+1)^3 >>}{ >>4 >>}\\ >>&= \dots \\ >>&= >>\frac{(k+1)^2(k+2)^2}{4}\\ >>&= >>\pa{\frac{(k+1)(k+2)}{2}}^2 \\ >>\end{align} >>$