Induction Proof with a $\neq$ 1

Question

For $a \neq 1$ and $n>0$, $(1-a^{n+1})/(1-a)=1+a+a^2+...+a^n$

How do you prove this by induction?

Answer 1

Assume $P(n): \dfrac{1-a^{n+1}}{1-a} = 1+a+a^2+...+a^n$ is true . We show $P(n+1)$ is true, i.e.

We have: $\dfrac{1-a^{n+2}}{1-a} = \dfrac{1-a^{n+1}}{1-a} + a^{n+1} = (1+a+...a^n) + a^{n+1}$, by the induction step.

We're done.

Answer 2

For $n=1$, $LHS=1+a=RHS$

Assume this is true for $n=p\geq 1$. Then $\displaystyle \frac{1-a^{p+1}}{1-a}=1+a+a^2+...+a^p$. Now add $a^{p+1}$ to both sides.

Answer 3

The statement holds for $n=1$.

We can rewrite the statement for $n$ as

$$1 - a^{n+1} = (1-a)(1+a+a^2+...+a^n).$$

Assuming this to be true, the case for $n+1$ is

$$1 - a^{n+2} = (1-a)(1+a+a^2+...+a^n + a^{n+1}) \\ = (1-a)(1+a+a^2+...+a^n) + (1-a)a^{n+1} \\ = [1 - a^{n+1}] + a^{n+1} - a^{n+2} = 1-a^{n+2},$$

and we've won.

(The bracketed term on the last line made use of the inductive assumption.)

Answer 4

let $P(n)$ represent the statement that ($a \ne 1$) $$ \frac{1-a^{n+1}}{1-a} = \sum_{k=0}^n a^k $$ then $P(0)$ is easily seen to be true. suppose the statement $P(n)$ were true. then adding $a^{n+1}$ to both sides we have $$ \frac{1-a^{n+1}}{1-a} +a^{n+1}= \sum_{k=0}^n a^k + a^{n+1} $$ i.e. $$ \frac{1-a^{n+1} +a^{n+1}(1-a)}{1-a} = \sum_{k=0}^{n+1} a^k $$ the LHS simplifies to $\frac{1-a^{n+2}}{1-a} $, so we have shown that $P(n) \Rightarrow P(n+1)$ which is the required inductive step to bootstrap the result through all the positive integers