Understanding Summation sequence

Question

I'm trying to wrap my head around this summation. I understand basic ones for the most part, stop, start, etc, but I don't understand this one in particular.

Answer 1

The summation starts from 0 and ends with n.

It is equal to $x^0+x^1+x^2+....+x^n$

$\implies 1+x^1+x^3+....+x^n$

which is nothing but finite geometric series whose summation can be given by

$\frac {a(1-x^{n+1})}{1-x}$ where $a=1$ the first term of the series

Answer 2

Notice that $$x^{n+1} - 1 = (x -1)(1 + x + ... + x^n) \Rightarrow \frac{x^{n+1} - 1}{x-1} = 1 + x + ... x^n = \sum_{k=0}^{n} x^k $$

Which can be easily shown by induction.

Answer 3

Let $$S=\sum_{k=0}^{n}x^k.$$ Then, we have $$xS=\color{red}{x^1+x^2+\cdots+x^{n-1}+x^n}+x^{n+1}$$ $$S=x^0+\color{red}{x^1+x^2+\cdots+x^{n-1}+x^n}.$$ Subtracting the latter from the former gives us $$xS-S=x^{n+1}-x^0.$$ Hence, we have $$\sum_{k=0}^{n}x^k=S=\frac{x^{n+1}-1}{x-1}$$ for $x\not=1$.

Answer 4

This is a standard trick for summing a geometric series. Put $$S_n=\sum_{i=0}^n x^i=1+x+\cdots+x^n$$ Then $$xS_n=x+x^2+\cdots+x^{n+1}$$ Then subtract these two to get $$\require{cancel}xS_n-S_n=(\cancel{x}+\cancel{x^2}+\cdots+x^{n+1})-(1+\cancel{x}+\cdots+\cancel{x^n})$$ $$(x-1)S_n =x^{n+1}-1$$ and as long as $x\neq 1$, the factor $(x-1)$ is nonzero so we can divide to get $$S_n =\boxed{\dfrac{x^{n+1}-1}{x-1}}$$

Note that this holds for any such finite sum, regardless of whether or not the infinite series converges.

In case $x=1$, the sum is simply $$S_n =\underbrace{1+1+\cdots+1}_{n \textrm{ copies}}= n$$