Vtyjkyuk

Let $P(n)$ be the statement
$$
P(n) : sumlimits^n_i=0r^i = dfrac1-r^1+n1-rtext for all n in mathbbNtext.
$$
I am stuck at the base case:
$$P(1):1 + r = dfrac1-r^21-rtext.$$

I am stuck as to how I can show $P(1)$ is true.

Note: First note that you are really summing the geometric series ($a,rinmathbbR, aneq 0, rneq 1$)
$$
a+ar+ar^2+cdots+ar^n=afracr^n+1-1r-1=afrac1-r^n+11-r
$$
with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.

Claim: Let $a$ and $r$ be real numbers with $aneq 0$ and $rneq 1$. Prove that for each integer $ngeq 1$ that
$$
a+ar+ar^2+cdots+ar^n=afracr^n+1-1r-1.
$$
Proof. Fix $rinmathbbR, rneq 1$, and let $S(n)$ denote the following statement for $ngeq 1$:
$$
S(n) : 1+r+r^2+cdots+r^n=fracr^n+1-1r-1.
$$
In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $ngeq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $rneq 0$ without loss of generality.

Base step ($n=1$): The statement $S(1)$ says that $1+r=fracr^2-1r-1$, which is true since $r^2-1=(r+1)(r-1)$ and $rneq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=fracr-1r-1$.]

Inductive step ($S(k)to S(k+1)$): Fix some $kgeq 1$ and suppose that $S(k)$ is true; that is, assume that
$$
S(k) : 1+r+r^2+cdots+r^k=fracr^k+1-1r-1
$$
is true. We must show that $S(k+1)$ follows where
$$
S(k+1) : 1+r+r^2+cdots+r^k+r^k+1=fracr^k+2-1r-1.
$$
Starting with the left-hand side of $S(k+1)$,
beginalign
textLHS &= colorred1+r+r^2+cdots+r^k+r^k+1tagby definition\[1em]
&= colorredfracr^k+1-1r-1+r^k+1tagby $S(k)$\[1em]
&= fracr^k+1-1r-1+fracr^k+1(r-1)r-1tagcommon denom.\[1em]
&= fracr^k+1-1+r^k+1(r-1)r-1tagcombine like terms\[1em]
&= fracr^k+1-1+r^k+2-r^k+1r-1tagexpand\[1em]
&= fracr^k+2-1r-1,tagsimplify
endalign
we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)to S(k+1)$, completing the inductive step.

By mathematical induction, for all $ngeq 1, S(n)$ holds true. $blacksquare$

Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $sum_i=0^n 2^i=2^n+1-1$. One can easily set $r=2$ in what we just proved to see that this result is true.

$$1+r=frac1-r^21-r$$should lead you to check if $$(1+r)(1-r)=1-r^2.$$
I don't need to tell you more.