Proof: (related to "Geometric Sum") 

• By hypothesis, $a,b\in R$ are elements of a unit ring $R,$ and $n\in\mathbb N$ is a natural number.
• We provide a proof by induction on $n.$
• Base case: $n=0$
  $$(a-b)\sum_{k=0}^0 a^k\cdot b^{-k}=(a-b)\cdot 1=a^1-b^1.$$
• Induction step $n\to n+1$
  • Assume, $$(a-b)\sum_{k=0}^n a^k\cdot b^{n-k}=a^{n+1}-b^{n+1}$$ holds for some $n\ge 0.$
  • Then $$\begin{align}(a-b)\sum_{k=0}^{n+1} a^k\cdot b^{n+1-k}&=(a-b)\sum_{k=0}^{n} a^k\cdot b^{n+1-k}+(a-b)\cdot ( a^{n+1}b^{0})\nonumber\\
    &=b(a-b)\sum_{k=0}^{n} a^k\cdot b^{n-k}+a^{n+2}-a^{n+1}b\nonumber\\
    &=b(a^{n+1}- b^{n+1})+a^{n+2}-a^{n+1}b\nonumber\\
    &=ba^{n+1}- b^{n+2}+a^{n+2}-a^{n+1}b\nonumber\\
    &=a^{n+2}-b^{n+2}\nonumber\\
    \end{align}$$
• Thus, the geometric sum holds for all $n\in\mathbb N.$

| | | | created: 2020-06-25 18:58:47 | modified: 2020-06-25 18:58:59 | by: | references: [8311]