\$egingroup\$ This is not true unless \$x=0\$. You have only a finite sum on the left, which is not a "series". \$endgroup\$
I am not sure if it is just a notational error, or a genuine misunderstanding, but the correct formula is\$\$1 + x + x^2 + x^3 + cdots + x^n + cdots = frac11-x\$\$Notice that unlike the question in the OP, the sum on the left does not stop at the \$x^n\$ term, but continues.

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As khổng lồ why this is true, there are a few different approaches of varying levels of formality. One (rather nonrigorous) approach is to start by calling the expression on the left \$S\$, and thinking about what happens if you multiply \$S\$ by \$x\$. We have\$\$S = 1 + x + x^2 + x^3 + cdots\$\$\$\$xS = x + x^2 + x^3 + cdots\$\$

Subtracting the bottom equation from the top, we see that everything on the right cancels except for the \$1\$ in the vị trí cao nhất equation; that is, we have\$\$S - xS = 1\$\$\$\$S(1-x) = 1\$\$\$\$S = frac11-x\$\$The reason I say this is a rather nonrigorous approach is that it begins by assuming that the infinite sum of the left-hand side converges lớn some number \$S\$. But how vị we know it converges at all? In fact, it doesn"t converge if \$|S| ge 1\$, & in that case this entire argument is bogus. Another problem with this approach is that it rather casually assumes that you can just cancel the sums of one infinite series off of the sums of the other; again, one really needs khổng lồ give careful consideration to issues of convergence in order to lớn justify this rigorously.

A second, slightly different way of approaching this is lớn consider the expression\$\$(1-x)(1 + x + x^2 + x^3 + cdots)\$\$Using the distributive property one gets\$\$(1 + x + x^2 + x^3 + cdots) - (x + x^2 + x^3 + cdots)\$\$and again everything cancels except the \$1\$ in the first pair of parentheses, so\$\$(1-x)(1 + x + x^2 + x^3 + cdots) = 1\$\$from which the desired conclusion follows.

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This second approach is really just a slightly rewritten version of the first approach, và suffers from the same problems.

A more careful approach needs lớn begin with the formula for a finite series. Call \$S_n\$ the sum of the first \$n\$ terms. Then

\$\$S_n = 1 + x + x^2 + x^3 + cdots + x^n = frac1-x^n+11-x\$\$which can be proven using a variation of the two approaches described above. Then we define \$1 + x + x^2 + cdots\$ khổng lồ mean \$lim_n o infty S_n\$. If \$|x|