How do I go about proving $\sum \limits _{j=0} ^{n-1} q^j = \tfrac {1 - q^n} {1-q}$ inductively? I can't seem to work with $n-1$.
So far, I have $\tfrac{1 - q^{n-1}} {1-q}+q^n$ = $\tfrac {1 - q^n} {1-q}$
$\tfrac{1 - q^{n-1}+q^n-q^{n+1}} {1-q}$ = $\tfrac {1 - q^n} {1-q}$
I can't simplify the numerator on the left side.
Is this correct?: $-q^{n+1}-q^{n-1}$ = $-2q^n$?
Then I will have: $1-2q^n$+$q^n$=$1-q^n$
That's because you don't want to show this:
$\tfrac{1 - q^{n-1}} {1-q}+q^n =\tfrac {1 - q^n} {1-q} $.
You want to show this:
$\tfrac{1 - q^{n-1}} {1-q}+q^{n-1} =\tfrac {1 - q^n} {1-q} $.
The exponent of $q$ in the fraction is one more than the exponent in the last term in the sum.
So if the sum goes up to $q^k$, the numerator of the fraction is $1-q^{k+1} $.
Here is a more explicit explanation.
You want to show that $\sum \limits _{j=0} ^{n-1} q^j = \tfrac {1 - q^n} {1-q} $.
Putting $n+1$ for $n$, this is $\sum \limits _{j=0} ^{n} q^j = \tfrac {1 - q^{n+1}} {1-q} $.
The difference of the two left hand sides is $\sum \limits _{j=0} ^{n} q^j-\sum \limits _{j=0} ^{n-1} q^j =q^n $.
The difference of the two right hand sides is
$\begin{array}\\ \tfrac {1 - q^{n+1}} {1-q}-\tfrac {1 - q^{n}} {1-q} &=\tfrac {(1 - q^{n+1})-(1 - q^{n})} {1-q}\\ &=\tfrac {q^n - q^{n+1}} {1-q}\\ &=\tfrac {q^n(1 - q)} {1-q}\\ &= q^n\\ \end{array} $
Since the two sides agree for $n=1$, and their differences are the same (both $q^n$) when you go from $n$ to $n+1$, the two sides are always the same.
This is a frequently useful method of proof by induction:
To show that $s(n) = t(n)$ for all $n$, show that $s(1) = t(1)$ and $s(n+1)-s(n) = t(n+1)-t(n) $.