Induction proof on $\sum_{i=0}^n m^i=\frac{m^{n+1}-1}{m-1}$

Question

For natural numbers $m\geq2$ and $n\geq1$, prove with induction that $\sum\limits_{i=0}^n m^i=\frac{m^{n+1}-1}{m-1}$.

My first intuition is to prove that $n+1,m$ is true, then $n,m+1$, then possibly $n+1,m+1$, though I'm uncertain whether the last is needed in general.

Proving $n+1,m$ holds was not difficult, but now I'm having trouble with $n,m+1$.

wts: $$\sum_{i=0}^n (m+1)^i=\frac{(m+1)^{n+1}-1}{(m+1)-1}=\frac{(m+1)^{n+1}-1}{m}$$

I expanded the summation, but that didn't help much to rearrange to the above.

@Dr.SonnhardGraubner Would induction not be required on $m$? (I did successfully use induction on $n$) — user154989, May 25 '17 at 20:46
Induction is not required on $m$ because both the basis (which is $\sum_0^1 x^i = 1+x = \frac{x^2-1}{x-1}$ is true for any real value of $m>1$ (which I therefore called $x$), and the induction step also works for arbitrary values of $m$ (or $x$). — Mark Fischler, May 25 '17 at 20:54

score 2 · Accepted Answer · answered May 25 '17 at 21:04

You should work on$n$ $$p_1 \to n=1 \to \sum_{i=0}^1 m^i=\frac{m^{1+1}-1}{m-1} \checkmark$$ $$p_k \to n=k \to \sum_{i=0}^k m^i=\frac{m^{k+1}-1}{m-1} \checkmark$$ $$p_{k+1} \to n=1 \to \sum_{i=0}^{k+1} m^i=\frac{m^{k+1+1}-1}{m-1} ?$$ form $p_k$,$p_{k+1}$ we have $$\sum_{i=0}^{k+1} m^i=\sum_{i=0}^{k} m^i+(m^{k+1})\\= \frac{m^{k+1}-1}{m-1}+(m^{k+1})=\\ \frac{m^{k+1}-1+(m-1)m^{k+1}}{m-1}=\\ \dfrac{m^{k+2}-1}{m-1}\checkmark$$

Induction proof on $\sum_{i=0}^n m^i=\frac{m^{n+1}-1}{m-1}$

1 Answers1