Intuition behind the finite geometric series formula?

Question

Can anyone give some intuition or insight on why $S_n = a(\frac{1-r^n}{1-r})$ works? (I've seen the proof but I like being able to visualize to think about formulas in different ways.)

Answer 1

You can visualize it by looking this identity as identity of polynomials: $$ (1-x)(1+x)=1-x^2 $$ $$ (1-x)(1+x+x^2)=1-x^3 $$ and in general $$ (1-x)(1+x+x^2+x^3+\ldots+x^{n-1})=1-x^n $$

Answer 2

Considering, when we rearrange the formula, we get:

Sn(r) - Sn = a(r^n) - a OR Sn - Sn(r) = a - a(r^n)

But what is Sn(r)?

Sn(r) = (r)[a + ar + ar^2 + ... + ar^(n-1)]

Hence, if there is an 'n' number of terms, then there is an 'n' number of terms which changes by a factor of 'r'. Equivalent to an 'n' number of 'r's:

(r)[a + ar + ar^2 + ... + ar^(n-1)] = ar + ar^2 + ar^3 + ... + a(r^n)

See? An 'n' number of 'r's. Makes much more sense now. Then, let's look at it this way:

Sn(r) - a(r^n) = Sn - a

Remember what Sn(r) is?

[ar + ar^2 + ar^3 + ... + a(r^n)] - ar^n = [a + ar + ar^2 + ... + ar^(n-1)] - a

ar + ar^2 + ... + ar^(n-1) = ar + ar^2 + ... + ar^(n-1)

Notice that this is somewhat similar to what "boaz" suggested. However, this is more general in the rearranged form, whereas "boaz"'s is a consideration in the simplified form.

For me, the algebraic expression is intuition enough, at least, at this level. However, I understand that for some people, they require concepts to be framed in their first language for greater intuition, which is not always easy as such languages are not mathematically and/or logically inclined.

The difference between the summation by a factor 'r' and the initial term by a factor 'r' is equal to the difference between the summation by a factor '1' and the initial term by a factor '1'. Because, the summation by a factor 'r' is such that every single term of the geometric series changes by a factor 'r', and for an 'n' number of terms, the initial term follows as in 'a(r^n)'. It is "corrected".