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Use the second principle of finite induction to establish that for all $n\geq1$ : $$a^n-1=(a-1)\left(a^{n-1}+a^{n-2}+a^{n-3}+\cdots+a+1\right) $$ Step by step explanation please! I'm confused how the second principle of finite induction is different from the first.

Martin Sleziak
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Lil
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2 Answers2

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You can use that \begin{align} a^{k+1}-1&=a(a^k-1)+(a-1)\\ &=a(a-1)(a^k+\cdots+a+1)+(a-1)\\ &=(a-1)(a^{k+1}+\cdots+a^2+a)+(a-1)\\ &=(a-1)(a^{k+1}+\cdots+a+1) \end{align} assuming you know that the theorem holds for $n=k$, but this is just simple induction (I think), not the so called 'second principle' of finite induction.

To answer your question:
The second principle uses all cases $n=1$ up to $n=k$ as induction hypothesis, instead of only $n=k$. See this answer on a similar question for more information.

EDIT
A quick google search returned the page above, which uses the similar wording. I had never heard of it (the phrasing 'second principle' instead of 'strong induction') myself too.

Community
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Ragnar
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Second principle of finite induction establishes if the given statement is true for $n$ from $1,2,3... n$. Then it is true for $n+1$.

$$a^1 -1= a-1 \\ a^2 -1= (a-1)(a+1)$$ say $n \ge 3$ assume $$a^{n} -1 =(a-1)(a^{n-1} +a^{n-2} +a^{n-3}.....+a+1)$$ we have to prove this for $n+1$. $a^{n+1} -1$ can be written as $(a+1)(a^{n-1})-a(a^{n-1} -1)$ we already know $a^n -1$ is $(a-1)(a^{n-1} +a^{n-2} +a^{n-3}.....+a+1)$

Let $S=a^{n-1} +a^{n-2} +a^{n-3}.....+a+1$ so I can write $a^{n+1} -1$ as $(a+1)(a-1)S -a(S-a^{n-1})(a-1)$ which on simplifying gives the required statement to prove

Hope it helps.

Shashaank
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