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If $r\neq 1$, show that $$a+ar+ar^2+\cdots+ar^n=\frac{a\left(r^{n+1}-1\right)}{r-1}$$ for any positive integer $n$

I seem to be doing something wrong could somebody tell me what is wrong with my method?

$n=1:$

$$ar^1=\frac{a(r^{1+1}-1)}{r-1}$$ $$ar=\frac{a(r-1)(r+1)}{r-1}$$ $$ar = a(r+1)$$ I can't see anything wrong my working, am i interpreting the question wrong?

lioness99a
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Sonny Da Silva-Peters
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    When $n=1$, it should be $a+ar^1$, as when $n=0$, the $L.H.S.$ should be $a$. – BAI Jun 12 '17 at 09:19
  • the question says prove for any positive interger of n, does that mean my final proof would read therefore true for all $n\ge1$ or would it conclude true for all $n\ge0$? – Sonny Da Silva-Peters Jun 12 '17 at 09:22
  • you could see that for $n=0$ the proposition still holds. However you could also try to minus $a$ from both sides. For these kinds of questions I recommend you to try to write it out in summation notation first. – BAI Jun 12 '17 at 09:24
  • okay thank you for your help – Sonny Da Silva-Peters Jun 12 '17 at 09:25
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    Step one: If $a = 0$ you are done, if not cancel it out to make things a lot easier. – Dirk Jun 12 '17 at 09:28

3 Answers3

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On the left hand side, it should be $a+ar$ instead of $ar$, since you're checking the case for $n=1$.

Lazy Lee
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For $n=1$ you have to show:

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

Fred
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Hint:

for $n=1$ the term is $a+ar^1=a(1+r)$

Emilio Novati
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