Induction Outset

Question

I am referring to this older question, which I found here for preparation:

Could somebody outline the Induction outset and the goal? I am unsure where to put in my (n + 1). My approach would be: 2⁰ - 1 + 2^(n+1) -1

score 0 · Accepted Answer · answered Nov 23 '20 at 15:35

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Base case $n=0$: $\sum\limits_{j=0}^0 2^j=2^{1}-1$.

Inductive step: assuming $\sum\limits_{j=0}^n 2^j=2^{n+1}-1$, show that $\sum\limits_{j=0}^{n\color{blue}{+1}} 2^j=2^{(n\color{blue}{+1})+1}-1$.

(Put $n\color{blue}{+1}$ wherever $n$ appeared.)

answered Nov 23 '20 at 15:35

J. W. Tanner

Alright, I got the very last part. This is my goal. But how do I start? So what does the sum sign on the left say? – Luise Nov 23 '20 at 15:50
Note that $\sum\limits_{j=0}^{n\color{blue}{+1}} 2^j=2^{n+1}+\sum\limits_{j=0}^{n } 2^j$ – J. W. Tanner Nov 23 '20 at 15:51
and where is the "-1" in this? – Luise Nov 23 '20 at 16:01
That would mean: 2^(n+1) + 2^(n+2) -1 ? – Luise Nov 23 '20 at 16:03
That would mean $2^{n+1}+2^{n+1}-1$, which is $2^{n+2}-1$, q.e.d. – J. W. Tanner Nov 23 '20 at 16:06
And a step in between would be 4^(n+1)-1 = 2^(n+2) - 1? – Luise Nov 23 '20 at 16:15
No, it would be $2^{n+1}+2^{n+1}= 2^{n+1}2=2^{n+1}2^1=2^{n+1+1}=2^{n+2}$ – J. W. Tanner Nov 23 '20 at 16:19

score 0 · Answer 2 · answered Nov 23 '20 at 16:00

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Generalise: $$1+x+x^2+\dots +x^n=\frac{x^{n+1}-1}{x-1}.$$ This formula is the simplest nontrivial model when one teaches proofs by induction.

The inductive step begins with \begin{align} 1+x+x^2+\dots+x^{n+1}&=(1+x+x^2+\dots +x^n)+x^{n+1}\\ &=\frac{x^{n+1}-1}{x-1}+x^{n+1}=\dots \end{align}

answered Nov 23 '20 at 16:00

Bernard

I would basically need the meaning of the sum sign as I am unsure about the power of j – Luise Nov 23 '20 at 16:04
@Luise: it means what's on the left side of the first equation of my answer, with $2$ in the place of $x$ – you add all powers of $2$, from $0$ to the $n$-th power. – Bernard Nov 23 '20 at 16:10

2 Answers2