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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.)

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    $\begingroup$ Everybody has his own way of thinking. Here you would get many different ideas about the same thing but what you have in your mind may be the best, so try to visualize it. The more ideas you would get here you would get confused more possibly. $\endgroup$ – Mayank Deora Jun 14 '16 at 11:20
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    $\begingroup$ Take a paper ,divide it into two equal parts, divide one of them into two equal parts ,keep on dividing n number of times..........................what you would get is beauty of math. $\endgroup$ – Mayank Deora Jun 14 '16 at 11:24
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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 $$ (1-x)(1+x+x^2+x^3+\ldots+x^{n-1})=1-x^n $$

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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".

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