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I have problem. Prove this using Mathematical Induction. I am a newbie in Mathematics. Please help me.

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

This is my way for get the proof

Basic Induction: $$ p(1)= a^1 = 1-a^1+1+1/1-a$$ $$ = 1-a^3/1-a$$

Really I don't understand this case.

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  • $\begingroup$ The sum of the first $n$ terms in a geometric series is given by $\dfrac{1-r^{n+1}}{1-r}$. Is this what you intended to write in the OP? You did not put proper braces or parenthesis around terms so it is very difficult to parse your work. $\endgroup$ – Cameron Williams Nov 12 '13 at 6:19
  • $\begingroup$ The basic induction should be for n=0, then $1 = \frac{1-a}{1-a} = 1$. Now assume it's true for n=k and prove it for n=k+1. Can you take it from here? $\endgroup$ – DanielY Nov 12 '13 at 6:27
  • $\begingroup$ For some intuition, note that $a = 10$, this equation tells us $999 = 1000 - 1$, etc. $\endgroup$ – littleO Nov 12 '13 at 6:50
  • $\begingroup$ See math.stackexchange.com/questions/11703/… $\endgroup$ – Martin Sleziak Nov 12 '13 at 17:22
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The basic induction should be for n=0, then $1 = \frac{1-a}{1-a} = 1$. Now assume it's true for $n=k$ and prove it for $n=k+1$.

So, $p(k+1) = 1+a+a^2+...+a^k +a^{k+1} = \frac{1-a^{k+1}}{1-a} + a^{k+1} = \frac{1-a^{k+1}+a^{k+1} - a^{k+2}}{1-a} = \frac{1-a^{k+2}}{1-a}$ which is exactly what we want

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    $\begingroup$ this is so cool, thanks for the answer .. $\endgroup$ – viyancs Nov 12 '13 at 6:32
  • $\begingroup$ you welcome :) induction is cool, I agree $\endgroup$ – DanielY Nov 12 '13 at 6:35
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This summation begins at $n=0$. If $n=0$, then we see that $a^0=1={1-a\over 1-a}={1-a^{0+1}\over 1-a}$. If $n=1$, then $1+a=(1+a){1-a\over 1-a}={1-a^2\over 1-a}$. Assume that $$1+a+a^2+\cdots+a^k={1-a^{k+1}\over 1-a}$$ is true for some arbitrary positive integer $k$. We must show that $$1+a+a^2+\cdots+a^k+a^{k+1}={1-a^{k+2}\over 1-a}.$$ Using our induction hypothesis consider $$(1+a+a^2+\cdots+a^k)+a^{k+1}={1-a^{k+1}\over 1-a}+a^{k+1}.$$ Rewrite the right-hand side to obtain $${1-a^{k+1}\over 1-a}+a^{k+1}={1-a^{k+1}+(1-a)a^{k+1}\over 1-a}$$ which becomes $${1-a^{k+1}+a^{k+1}-a^{k+2}\over 1-a}={1-a^{k+2}\over 1-a}.$$ Thus by the Principle of Mathematical Induction $$1+a+a^2+\cdots+a^n={1-a^{n+1}\over 1-a}$$ for all $n\geq0$.

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  • $\begingroup$ this answer more details thanks you :) $\endgroup$ – viyancs Nov 13 '13 at 3:52
  • $\begingroup$ @viyancs You're welcome :) $\endgroup$ – 1233dfv Nov 13 '13 at 12:47
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For those who don't like induction:

Let $S_n = \sum_{i=0}^{n}a^i$, then $aS_n = \sum_{i=1}^{n+1}a^i$.

So $(1-a)S_n = 1 - a^{n+1}$, the conclusion follows.

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Your error is $$p(1)=a^1$$ The correct statement should be $$p(1)=1+a^1=1+a$$ $$=\frac{1-a^2}{1-a}$$

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