Prove that $1+a+a^2+\cdots+a^n=(1-a^{n+1})/(1-a)$. 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.
 A: 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 $n\geq0$.
A: 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.
A: Your error is $$p(1)=a^1$$ The correct statement should be $$p(1)=1+a^1=1+a$$ $$=\frac{1-a^2}{1-a}$$
A: 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
