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Problem: Use the Second Principle of Finite Induction to establish that for all $n\ge1$ $a^n-1 = (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1)$. [Hint: $a^{n+1}-1=(a+1)(a^n-1)-a(a^{n-1}-1)$]. --- Scratch work: --- Base Cases: $a^1-1=(a-1)$ $a^2-1=(a-1)(a+1)$ --- Inductive Step: Let $k\ge3$ for some integer $k$. We'll then assume that the following are true: $a^k-1=(k-1)(\sum_{k=0}^{k-1}a^k)$ and $a^{k-1} -1 = (a-1)(\sum_{k=0}^{k-2}a^k)$ It then holds, via the hint, that: $a^{k+1}-1 = (a+1)(a^k-1)-a(a^{k-1}-1)$ $= (a+1)[(a-1)(\sum_{k=0}^{k-1}a^k)] - a[(a-1)(\sum_{k=0}^{k-2}a^k)]$ $= (a-1)[(a+1)(\sum_{k=0}^{k-1}a^k) - a(\sum_{k=0}^{k-2}a^k)]$ $= (a-1)[a(\sum_{k=0}^{k-1}a^k) + \sum_{k=0}^{k-1}a^k - a(\sum_{k=0}^{k-2}a^k])$ $= (a-1)[\sum_{k=1}^{k}a^k + \sum_{k=0}^{k-1}a^k - \sum_{k=1}^{k-1}a^k]$ $= (a-1)[\sum_{k=1}^{k}a^k - \sum_{k=1}^{k-1}a^k + \sum_{k=0}^{k-1}a^k]$ $= (a-1)[a^k + \sum_{k=0}^{k-1}a^k]$ $= (a-1)[\sum_{k=0}^{k}a^k]$.
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Sun, 31 Jan 2021 14:05 GMT