I am using this formula when calculating geometric sequence sum, starting from $i = 0$:
$$\sum_{i = 0}^{n} ax^i = a\frac{1 - x^{n+1}}{1-x}$$
But what if I was starting from $i = 1$ instead?
Is this formula valid and will always work when $x \neq 1$?
$$\sum_{i=1}^{n}ax^{i} = a \frac{1-x^{n}}{1-x}$$
What if I was starting at some $i = m \quad \land \quad m < n$?
On
Hint: for any given $m$, $\sum_{i=m}^nax^i=\sum_{i=0}^nax^i-\sum_{i=0}^{m-1}ax^i$
More generally, suppose you have a sequence $p_i$ of numbers, and some "nice formula" $P(n)$, depending only on $n$, for $\sum_{i=0}^np_i=P(n)$.
Then we have a "nice formula" $P(n,m)$ for $\sum_{i=m}^n p_i$, namely, $P(n,m)=P(n)-P(m-1)$.
You can factorize $x^m$ out of the sum.
$$x^m+x^{m+1}+\cdots+x^n=x^m(1+x+x^2+\cdots+x^{n-m})$$
Thus you get $\quad\displaystyle\sum\limits_{i=m}^n x^i=x^m\sum\limits_{i=0}^{n-m} x^i=x^m\cdot\dfrac{1-x^{n-m+1}}{1-x}=\dfrac{x^m-x^{n+1}}{1-x}$
Multiplying by the const coefficient $a$, changes nothing of course, I omitted it to ease the understanding.