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How can I use the method of induction to show for any real number $r$ does not equal $1$ and any positive integer $n$

show that

$$1+r+r^2+\cdots+r^n=\frac{1r^{n+1}-1}{r-1}$$

for $n=1$ it seems to work

$$1+r+\cdots+r^n=(1+r)$$

then $\dfrac{r^2-1}{r-1}$ for the right side

$$\frac{(r-1)(r+1)}{r-1}=1+r$$

Thus the formula is true for $n=1$ then assume the formula is true $n=k$ ,$k$ is an integer greater than $1$.

$$1+r+\cdots+r^k=\frac{r^{k+1}-1}{r-1}$$

then

$$1+r+\cdots+r^k+r^{k+1}$$

then

$$\frac{1r^{k+1}-1}{r-1} + \frac{r^{k+1}}{1}$$

$$\frac{r^{k+1}+r-1}{r-1}=1+r+\cdots+r^k+r^{k+1}$$

would this be correct method of induction?

Michael Hardy
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Fernando Martinez
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2 Answers2

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Assuming $r\ne1$

Let $\displaystyle F(n): 1+r+r^2+....+r^n=\frac{r^{n+1}-1}{r-1}$ holds true for $n=m $

$\displaystyle\implies 1+r+r^2+....+r^m=\frac{r^{m+1}-1}{r-1}$

$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{m+1}-1}{r-1}+r^{m+1}=\frac{r^{m+1}-1+r^{m+2}-r^{m+1}}{r-1}$

$\displaystyle \implies 1+r+r^2+....+r^m+r^{m+1}=\frac{r^{(m+1)+1}-1}{r-1}$

So, $f(n)$ will hold true for $n=m+1$ if it holds true for $n=m$

Establish the base case $n=1$

lab bhattacharjee
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  • I see I think I get it the only thing is do you get $r^{m+2}-r^{m+1}$ on the numerator when you do $r^{m+1}(r-1)$ when you get the common denominator in the fraction – Fernando Martinez Dec 08 '13 at 19:08
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Again, you don't need induction, use perturbation for $S_n = \sum_{k=0}^{n}r^k$: $$ S_n = \sum_{k=0}^{n}r^k\\ S_n +r^{n+1}= \sum_{k=0}^{n}r^k +r^{n+1}\\ S_n +r^{n+1} = \sum_{k=0}^{n+1}r^k=r \sum_{k=0}^{n}r^k +1\\ \text{a bit of algebra}\\ S_n=\frac{1-r^{n+1}}{1-r} $$

Alex
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