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I have a question about methematical induction. I have to proof that:

$\require{cancel}$ $$\sum_{k=0}^n 11^k$$

Knowing that: $$\sum_{k=0}^n a^k = \frac{1 - a^{k+1}}{1-a} $$

I have the base:

$$ 11^0 = \frac{1-11^{0+1}}{1-11}$$ $$ 1 = \frac{-10}{-10} $$ $$ 1 = 1 $$

Hypothesis: $$ 1 + 11 + 121 + ... + 11^k = \frac{1-11^{k+1}}{1-11}$$

Conclusion: $$1 + 11 + 121 + ... + 11^k + 11^{k+1}= \frac{1-11^{k+2}}{1-11}$$ $$\frac{1-11^{k+1}}{1-11} + 11^{k+1}= \frac{1-11^{k+2}}{1-11}$$ $$\frac{1-11^{k+1}}{-10} + 11^{k+1}= \frac{1-11^{k+2}}{-10}$$

$$\frac{1-11^{k+1} + 10.11^{k+1}}{\cancel{-10}} = \frac{1-11^{k+2}}{\cancel{-10}}$$

Then: $$1-11^{k+1} + 10.11^{k+1} = 1-11^{k+2}$$

From here I don't know how to continue. I think I'm missing some properties. Any help will be welcomed.

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    $\begingroup$ I think you forgot to add what it should equal to. $\endgroup$ – Nescrio May 26 '16 at 19:59
  • $\begingroup$ What are you trying to prove? $\endgroup$ – Ben Sheller May 26 '16 at 20:04
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prove by induction

$\sum_\limits {k=0}^n 11^k = \frac{1-11^{k+1}}{1-11}$

You have already take care of the base case.

Inductive hypothesis:

suppose, $\sum_\limits {k=0}^n 11^k = \frac{1-11^{n+1}}{1-11}$

show that, $\sum_\limits {k=0}^{n+1} 11^k = \frac{1-11^{n+2}}{1-11}$

$\sum_\limits {k=0}^{n+1} 11^k\\ \sum_\limits {k=0}^{n} 11^k + 11^{n+1}$

$\frac{1-11^{n+1}}{1-11} + 11^{n+1}$ ,by the inductive hypothesis

$\frac{1-11^{n+1}+(1-11)11^{n+1}}{1-11}\\ \frac{1-11^{n+1}+11^{n+1}-11^{n+2}}{1-11}\\ \frac{1-11^{n+2}}{1-11}$

QED

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Directly from the inductive hypothesis for $k$ you have $$1+11+11^2+\dots+11^{k+1}=(1+11+11^2+\dots+11^k)+11^{k+1}=\left(\frac{1-11^{k+1}}{1-11}\right)+11^{k+1}$$ you should then continue $$=\frac{1-11^{k+1}+11^{k+1}(1-11)}{1-11}$$ $$=\frac{1-11^{k+2}}{1-11}$$ which is the required result for $k+1$, thus completing the induction.

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As far as I know, $$ \forall |x| > 0, \sum_{k=0}^{n} a^k = \frac{a^{n+1} - 1}{a-1}$$. Also, I think you forgot to mention what the summation should equal to.

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