Can I write $\sum_{n=1}^N x^n$ as $\sum_{n=1}^N x\cdot x^{n-1}$ and then apply the geometric series formula for sum of first $n$ terms with $a=x$?

Question

If the answer is yes, then what about the fact that $$\sum_{n=1}^N x^n$$ has $a=1, r=x$, which implies a sum of $\frac{1-x^N}{1-x}$ whereas $\sum_{n=1}^N x \cdot x^{n-1}$ has a sum of $$\frac{x(1-x^{N-1})}{1-x} = \frac{x-x^N}{1-x}$$ The numerators are different but the denominators are same, so the two formulas have different sums?

I think I'm missing something dumb here, like a division or multiplication by zeros somewhere, or something.

Also, assume $\lvert x \rvert <1$

Thanks.

Edit: My formulas in this post are incorrect. See comments below for correct ones, and explanation of why mine are incorrect.

Answer 1

Let $a_n$ be geometric series and $S_n$ the sum of first $n$ terms, thus $\displaystyle S_n=\frac{a_1(1-q^n)}{1-q}$.

In your case we have$$\sum_{n=1}^N x^n=x+x^2+\dots+x^N=\frac{x(1-x^N)}{1-x}$$ where $a_1=x$ and there are $n$ terms. $$\sum_{n=1}^{N}x^n=x\sum_{n=1}^{N}x^{n-1}=x\left(1+x+\dots+x^{N-1}\right)=x\cdot{\frac{1(1-x^N)}{1-x}}$$where $a_1=1$ and there are $n$ terms.

Easy to understand that both sums are equal

Answer 2

The basic formula is

$$\sum_{n=0}^N x^n=\frac{1-x^{N+1}}{1-x},$$ so that

$$\sum_{n=1}^N x^n=\frac{1-x^{N+1}}{1-x}-1$$ (first term missing) and $$\sum_{n=1}^N x\cdot x^{n-1}=x\sum_{n=0}^{N-1} x^{n}=x\frac{1-x^N}{1-x}$$ (by shifting of the index).

You can verify that both expressions equal

$$\frac{x-x^{N+1}}{1-x}.$$

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For instance, $$3+9+27+81=3(1+3+9+27)=\frac{2-243}{1-3}.$$

Answer 3

WLOG, $N=4$.

$$S:=x+x^2+x^3+x^4=x(1+x+x^2+x^3)=x(1+S-x^4)=xS+x-x^5,$$

then

$$S=\frac{x-x^5}{1-x}.$$