How to derive the formula for a geometric series sum

Any geometric series is defined by an initial term a, and a common ratio b

This means that we start with a and multiply by b to get each next term.

And so the general geometric series is written:

a*r^0,a*r^1,a*r^2,a*r^3,...,a*r^n ; where n+1 = no. of terms.

Now the sum of the above, S is:

S = a*r^0+a*r^1+a*r^2+a*r^3+...+ar^n

Factorise r out in most terms in right hand side:

S = a*r^0+r*(a*r^0+a*r^1+a*r^2+...+a*r^(n-1))

Now note that the sum within the brackets includes all terms in our original sum S, save the last term a*r^n. This means we can substitute (S-a*r^n) for it. Therefore,

S = a*r^0+r*(S-a*r^n)

Now all we are left to do is make the value of the sum, S, the subject of the equation:

S = a+r*S-a*r^(n+1)

S-r*S = a-a*r^(n+1)

S*(1-r) = a*(1-r*(n+1)); factorize S on LHS, and a on RHS.

S = a*(1-r*(n+1))/(1-r)


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